Eigenvalues of $2$ symmetric $4times 4$ matrices: why is one negative of the other?If the eigenvalues are...

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Eigenvalues of $2$ symmetric $4times 4$ matrices: why is one negative of the other?


If the eigenvalues are distinct then the eigenspaces are all one dimensionalCongruence of invertible skew symmetric matricesEigenvalues of a general block hermitian matrixEigenvalues of Overlapping block diagonal matricesHow to find a symmetric matrix that transforms one ellipsoid to another?The matrix of an endomorphismA conjecture regarding the eigenvalues of real symmetric matricesProve that the span of ${M_1, M_2, M_3}$ is the set of all symmetric $2times2$ matrices.Looking for properties of, or formulae for eigenvalues of a symmetric matrix reminiscent of Toeplitz matricesDo hermitian matrices commute when they occupy they same elements but have different values?













2












$begingroup$


Consider the following symmetric matrix:



$$
M_0 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



and a very similar matrix:



$$
M_1 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?



I also tried playing around with the values a little; for example, if the center block is $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ instead, then they do not share the same eigenvalues.





Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.



$$
M_2 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & e^{ix} & 3 \
2 & e^{-ix} & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.



Thanks.










share|cite|improve this question











$endgroup$












  • $begingroup$
    It's because of all the conveniently placed zeroes.
    $endgroup$
    – M. Vinay
    32 mins ago










  • $begingroup$
    @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
    $endgroup$
    – Troy
    30 mins ago






  • 1




    $begingroup$
    In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
    $endgroup$
    – M. Vinay
    16 mins ago






  • 2




    $begingroup$
    In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
    $endgroup$
    – leonbloy
    15 mins ago










  • $begingroup$
    @leonbloy that certainly narrows down the search for me, thanks for the input!
    $endgroup$
    – Troy
    13 mins ago
















2












$begingroup$


Consider the following symmetric matrix:



$$
M_0 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



and a very similar matrix:



$$
M_1 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?



I also tried playing around with the values a little; for example, if the center block is $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ instead, then they do not share the same eigenvalues.





Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.



$$
M_2 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & e^{ix} & 3 \
2 & e^{-ix} & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.



Thanks.










share|cite|improve this question











$endgroup$












  • $begingroup$
    It's because of all the conveniently placed zeroes.
    $endgroup$
    – M. Vinay
    32 mins ago










  • $begingroup$
    @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
    $endgroup$
    – Troy
    30 mins ago






  • 1




    $begingroup$
    In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
    $endgroup$
    – M. Vinay
    16 mins ago






  • 2




    $begingroup$
    In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
    $endgroup$
    – leonbloy
    15 mins ago










  • $begingroup$
    @leonbloy that certainly narrows down the search for me, thanks for the input!
    $endgroup$
    – Troy
    13 mins ago














2












2








2


1



$begingroup$


Consider the following symmetric matrix:



$$
M_0 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



and a very similar matrix:



$$
M_1 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?



I also tried playing around with the values a little; for example, if the center block is $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ instead, then they do not share the same eigenvalues.





Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.



$$
M_2 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & e^{ix} & 3 \
2 & e^{-ix} & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.



Thanks.










share|cite|improve this question











$endgroup$




Consider the following symmetric matrix:



$$
M_0 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



and a very similar matrix:



$$
M_1 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?



I also tried playing around with the values a little; for example, if the center block is $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ instead, then they do not share the same eigenvalues.





Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.



$$
M_2 =
begin{pmatrix}
0 & 1 & 2 & 0 \
1 & 0 & e^{ix} & 3 \
2 & e^{-ix} & 0 & 1 \
0 & 3 & 1 & 0
end{pmatrix}
$$



ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.



Thanks.







linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 mins ago









YuiTo Cheng

2,2734937




2,2734937










asked 1 hour ago









TroyTroy

4231519




4231519












  • $begingroup$
    It's because of all the conveniently placed zeroes.
    $endgroup$
    – M. Vinay
    32 mins ago










  • $begingroup$
    @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
    $endgroup$
    – Troy
    30 mins ago






  • 1




    $begingroup$
    In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
    $endgroup$
    – M. Vinay
    16 mins ago






  • 2




    $begingroup$
    In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
    $endgroup$
    – leonbloy
    15 mins ago










  • $begingroup$
    @leonbloy that certainly narrows down the search for me, thanks for the input!
    $endgroup$
    – Troy
    13 mins ago


















  • $begingroup$
    It's because of all the conveniently placed zeroes.
    $endgroup$
    – M. Vinay
    32 mins ago










  • $begingroup$
    @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
    $endgroup$
    – Troy
    30 mins ago






  • 1




    $begingroup$
    In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
    $endgroup$
    – M. Vinay
    16 mins ago






  • 2




    $begingroup$
    In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
    $endgroup$
    – leonbloy
    15 mins ago










  • $begingroup$
    @leonbloy that certainly narrows down the search for me, thanks for the input!
    $endgroup$
    – Troy
    13 mins ago
















$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
32 mins ago




$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
32 mins ago












$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
30 mins ago




$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
30 mins ago




1




1




$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
16 mins ago




$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
16 mins ago




2




2




$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
15 mins ago




$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
15 mins ago












$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
13 mins ago




$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
13 mins ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$

and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$

are conjugate, for precisely the same reason.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    7 mins ago










  • $begingroup$
    okay, this is amazing..
    $endgroup$
    – Troy
    1 min ago



















2












$begingroup$

This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}

And the cases of the third and fourth rows are obviously similar.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    16 mins ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    13 mins ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    11 mins ago



















1












$begingroup$

I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}

This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}

A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    20 mins ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    17 mins ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    13 mins ago












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$

and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$

are conjugate, for precisely the same reason.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    7 mins ago










  • $begingroup$
    okay, this is amazing..
    $endgroup$
    – Troy
    1 min ago
















4












$begingroup$

$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$

and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$

are conjugate, for precisely the same reason.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    7 mins ago










  • $begingroup$
    okay, this is amazing..
    $endgroup$
    – Troy
    1 min ago














4












4








4





$begingroup$

$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$

and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$

are conjugate, for precisely the same reason.






share|cite|improve this answer









$endgroup$



$$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\
a_{21}&a_{22}&a_{23}&a_{24}\
a_{31}&a_{32}&a_{33}&a_{34}\
a_{41}&a_{42}&a_{43}&a_{44}}$$

and
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$

are conjugate, for precisely the same reason.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 10 mins ago









Lord Shark the UnknownLord Shark the Unknown

108k1162135




108k1162135








  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    7 mins ago










  • $begingroup$
    okay, this is amazing..
    $endgroup$
    – Troy
    1 min ago














  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    7 mins ago










  • $begingroup$
    okay, this is amazing..
    $endgroup$
    – Troy
    1 min ago








1




1




$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
7 mins ago




$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
7 mins ago












$begingroup$
okay, this is amazing..
$endgroup$
– Troy
1 min ago




$begingroup$
okay, this is amazing..
$endgroup$
– Troy
1 min ago











2












$begingroup$

This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}

And the cases of the third and fourth rows are obviously similar.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    16 mins ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    13 mins ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    11 mins ago
















2












$begingroup$

This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}

And the cases of the third and fourth rows are obviously similar.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    16 mins ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    13 mins ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    11 mins ago














2












2








2





$begingroup$

This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}

And the cases of the third and fourth rows are obviously similar.






share|cite|improve this answer











$endgroup$



This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
begin{align*}
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
end{align*}

And the cases of the third and fourth rows are obviously similar.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 mins ago

























answered 19 mins ago









M. VinayM. Vinay

7,33322136




7,33322136












  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    16 mins ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    13 mins ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    11 mins ago


















  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    16 mins ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    13 mins ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    11 mins ago
















$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
16 mins ago




$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
16 mins ago












$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
13 mins ago




$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
13 mins ago












$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
11 mins ago




$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
11 mins ago











1












$begingroup$

I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}

This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}

A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    20 mins ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    17 mins ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    13 mins ago
















1












$begingroup$

I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}

This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}

A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    20 mins ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    17 mins ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    13 mins ago














1












1








1





$begingroup$

I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}

This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}

A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.






share|cite|improve this answer











$endgroup$



I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

The characteristic polynomials of $M_a$ and $M_{-a}$ are
begin{align*}
chi_{M_a}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) t^{2} + 10 , a t + 25
end{align*}

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
begin{align*}
0
&= chi_{M_a}(t) \
&= {lambda}^{4} - {left(a^{2} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-lambda)^{2} + 10 , a (-lambda) + 25 \
&= chi_{M_{-a}}(-lambda)
end{align*}

This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=left[begin{array}{rrrr}
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
end{array}right]
$$

In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
begin{align*}
chi_{M_{a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) t^{2} + 10 , a t + 25
end{align*}

A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 18 mins ago

























answered 28 mins ago









Brian FitzpatrickBrian Fitzpatrick

21.8k42959




21.8k42959












  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    20 mins ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    17 mins ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    13 mins ago


















  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    20 mins ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    17 mins ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    13 mins ago
















$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
20 mins ago




$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
20 mins ago












$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
17 mins ago




$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
17 mins ago












$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
13 mins ago




$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
13 mins ago


















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