Eigenvalues of special singular matrix
$begingroup$
Let's consider a matrix $Ainmathbb{R}^{ntimes n}$, with eigenvalues $lambda_i$. We assume $A$ is positive definite, so $lambda_i>0$. Now lets consider the matrix
$$ B =lambda I - A,qquad lambda in lambda_i.$$
We know that $B$ is singular, even better we know $lambda_i$ are intended to be found by computing for which $lambda det(B)=0$. Now let us denote $mu_i$ to be the eigenvalues of $B$. We know that $mu_1=0$, but I was wondering if we can say anyting about the rest of the eigenvalues of B.
I considered the following example:
$$A = begin{pmatrix}14 & 38 & 26\
38 & 110 &94\
26 & 94 & 145end{pmatrix}$$
With eigenvalues $lambda = [0.1879, 36.6743, 232.1378] $. Then if we choos $B = lambda_1I-A$, we get that
$$ mu = [0, -36.6743, -232.1378] = lambda_1-lambda. $$
Now my question is if A) the statement $mu = lambda_i -lambda$ holds for any matrix $A$ and all eigenvalues $lambda_i$ and B) if this only holds for eigenvalues $lambda_i$, or holds for all matrices structured as $B = cI-A$, where $c$ can be any scalar value and this is actually a known property.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 21 '18 at 12:01
User123456789User123456789
512315
$endgroup$
add a comment |
$begingroup$
Let's consider a matrix $Ainmathbb{R}^{ntimes n}$, with eigenvalues $lambda_i$. We assume $A$ is positive definite, so $lambda_i>0$. Now lets consider the matrix
$$ B =lambda I - A,qquad lambda in lambda_i.$$
We know that $B$ is singular, even better we know $lambda_i$ are intended to be found by computing for which $lambda det(B)=0$. Now let us denote $mu_i$ to be the eigenvalues of $B$. We know that $mu_1=0$, but I was wondering if we can say anyting about the rest of the eigenvalues of B.
I considered the following example:
$$A = begin{pmatrix}14 & 38 & 26\
38 & 110 &94\
26 & 94 & 145end{pmatrix}$$
With eigenvalues $lambda = [0.1879, 36.6743, 232.1378] $. Then if we choos $B = lambda_1I-A$, we get that
$$ mu = [0, -36.6743, -232.1378] = lambda_1-lambda. $$
Now my question is if A) the statement $mu = lambda_i -lambda$ holds for any matrix $A$ and all eigenvalues $lambda_i$ and B) if this only holds for eigenvalues $lambda_i$, or holds for all matrices structured as $B = cI-A$, where $c$ can be any scalar value and this is actually a known property.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 21 '18 at 12:01
User123456789User123456789
512315
$endgroup$
add a comment |
$begingroup$
Let's consider a matrix $Ainmathbb{R}^{ntimes n}$, with eigenvalues $lambda_i$. We assume $A$ is positive definite, so $lambda_i>0$. Now lets consider the matrix
$$ B =lambda I - A,qquad lambda in lambda_i.$$
We know that $B$ is singular, even better we know $lambda_i$ are intended to be found by computing for which $lambda det(B)=0$. Now let us denote $mu_i$ to be the eigenvalues of $B$. We know that $mu_1=0$, but I was wondering if we can say anyting about the rest of the eigenvalues of B.
I considered the following example:
$$A = begin{pmatrix}14 & 38 & 26\
38 & 110 &94\
26 & 94 & 145end{pmatrix}$$
With eigenvalues $lambda = [0.1879, 36.6743, 232.1378] $. Then if we choos $B = lambda_1I-A$, we get that
$$ mu = [0, -36.6743, -232.1378] = lambda_1-lambda. $$
Now my question is if A) the statement $mu = lambda_i -lambda$ holds for any matrix $A$ and all eigenvalues $lambda_i$ and B) if this only holds for eigenvalues $lambda_i$, or holds for all matrices structured as $B = cI-A$, where $c$ can be any scalar value and this is actually a known property.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 21 '18 at 12:01
User123456789User123456789
512315
$endgroup$
Let's consider a matrix $Ainmathbb{R}^{ntimes n}$, with eigenvalues $lambda_i$. We assume $A$ is positive definite, so $lambda_i>0$. Now lets consider the matrix
$$ B =lambda I - A,qquad lambda in lambda_i.$$
We know that $B$ is singular, even better we know $lambda_i$ are intended to be found by computing for which $lambda det(B)=0$. Now let us denote $mu_i$ to be the eigenvalues of $B$. We know that $mu_1=0$, but I was wondering if we can say anyting about the rest of the eigenvalues of B.
I considered the following example:
$$A = begin{pmatrix}14 & 38 & 26\
38 & 110 &94\
26 & 94 & 145end{pmatrix}$$
With eigenvalues $lambda = [0.1879, 36.6743, 232.1378] $. Then if we choos $B = lambda_1I-A$, we get that
$$ mu = [0, -36.6743, -232.1378] = lambda_1-lambda. $$
Now my question is if A) the statement $mu = lambda_i -lambda$ holds for any matrix $A$ and all eigenvalues $lambda_i$ and B) if this only holds for eigenvalues $lambda_i$, or holds for all matrices structured as $B = cI-A$, where $c$ can be any scalar value and this is actually a known property.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 21 '18 at 12:01
User123456789User123456789
512315
asked Dec 21 '18 at 12:01
User123456789User123456789
512315
asked Dec 21 '18 at 12:01
User123456789User123456789
512315
asked Dec 21 '18 at 12:01
User123456789User123456789
512315
asked Dec 21 '18 at 12:01
User123456789User123456789
512315
512315
add a comment |
add a comment |
1 Answer
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$begingroup$
Suppose that $mu$ is an eigenvalue of $B$ and that $B=coperatorname{Id}-A$ for some matrix $A$ and some number $c$. Then there is a non-zero vector $v$ such that $B.v=mu v$. Butbegin{align}B.v=mu v&iff(coperatorname{Id}-A).v=mu v\&iff cv-A.v=mu v\&iff A.v=cv-mu v\&iff A.v=(c-mu)v.end{align}So, whenever $mu$ is an eigenvalue of $B$, $c-mu$ is an eigenvalue of $A$. By the same argument, if $lambda$ is an eigenvalue of $A$, $c-lambda$ is an eigenvalue of $B$. Note that the only thing I need for this to work is that $B$ is a square matrix. Being symmetric is not relevant.
edited Dec 21 '18 at 17:39
StubbornAtom
6,37831440
answered Dec 21 '18 at 12:10
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
$begingroup$
Thanks! Putting it like this I feel a bit stupid that I did not notice this myself...
$endgroup$
– User123456789
Dec 21 '18 at 12:17
1
$begingroup$
Believe me, I know that feeling. I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 21 '18 at 12:19
add a comment |
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1 Answer
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$begingroup$
Suppose that $mu$ is an eigenvalue of $B$ and that $B=coperatorname{Id}-A$ for some matrix $A$ and some number $c$. Then there is a non-zero vector $v$ such that $B.v=mu v$. Butbegin{align}B.v=mu v&iff(coperatorname{Id}-A).v=mu v\&iff cv-A.v=mu v\&iff A.v=cv-mu v\&iff A.v=(c-mu)v.end{align}So, whenever $mu$ is an eigenvalue of $B$, $c-mu$ is an eigenvalue of $A$. By the same argument, if $lambda$ is an eigenvalue of $A$, $c-lambda$ is an eigenvalue of $B$. Note that the only thing I need for this to work is that $B$ is a square matrix. Being symmetric is not relevant.
edited Dec 21 '18 at 17:39
StubbornAtom
6,37831440
answered Dec 21 '18 at 12:10
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
$begingroup$
Thanks! Putting it like this I feel a bit stupid that I did not notice this myself...
$endgroup$
– User123456789
Dec 21 '18 at 12:17
1
$begingroup$
Believe me, I know that feeling. I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 21 '18 at 12:19
add a comment |
$begingroup$
Suppose that $mu$ is an eigenvalue of $B$ and that $B=coperatorname{Id}-A$ for some matrix $A$ and some number $c$. Then there is a non-zero vector $v$ such that $B.v=mu v$. Butbegin{align}B.v=mu v&iff(coperatorname{Id}-A).v=mu v\&iff cv-A.v=mu v\&iff A.v=cv-mu v\&iff A.v=(c-mu)v.end{align}So, whenever $mu$ is an eigenvalue of $B$, $c-mu$ is an eigenvalue of $A$. By the same argument, if $lambda$ is an eigenvalue of $A$, $c-lambda$ is an eigenvalue of $B$. Note that the only thing I need for this to work is that $B$ is a square matrix. Being symmetric is not relevant.
edited Dec 21 '18 at 17:39
StubbornAtom
6,37831440
answered Dec 21 '18 at 12:10
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
$begingroup$
Thanks! Putting it like this I feel a bit stupid that I did not notice this myself...
$endgroup$
– User123456789
Dec 21 '18 at 12:17
1
$begingroup$
Believe me, I know that feeling. I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 21 '18 at 12:19
add a comment |
$begingroup$
Suppose that $mu$ is an eigenvalue of $B$ and that $B=coperatorname{Id}-A$ for some matrix $A$ and some number $c$. Then there is a non-zero vector $v$ such that $B.v=mu v$. Butbegin{align}B.v=mu v&iff(coperatorname{Id}-A).v=mu v\&iff cv-A.v=mu v\&iff A.v=cv-mu v\&iff A.v=(c-mu)v.end{align}So, whenever $mu$ is an eigenvalue of $B$, $c-mu$ is an eigenvalue of $A$. By the same argument, if $lambda$ is an eigenvalue of $A$, $c-lambda$ is an eigenvalue of $B$. Note that the only thing I need for this to work is that $B$ is a square matrix. Being symmetric is not relevant.
edited Dec 21 '18 at 17:39
StubbornAtom
6,37831440
answered Dec 21 '18 at 12:10
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
Suppose that $mu$ is an eigenvalue of $B$ and that $B=coperatorname{Id}-A$ for some matrix $A$ and some number $c$. Then there is a non-zero vector $v$ such that $B.v=mu v$. Butbegin{align}B.v=mu v&iff(coperatorname{Id}-A).v=mu v\&iff cv-A.v=mu v\&iff A.v=cv-mu v\&iff A.v=(c-mu)v.end{align}So, whenever $mu$ is an eigenvalue of $B$, $c-mu$ is an eigenvalue of $A$. By the same argument, if $lambda$ is an eigenvalue of $A$, $c-lambda$ is an eigenvalue of $B$. Note that the only thing I need for this to work is that $B$ is a square matrix. Being symmetric is not relevant.
edited Dec 21 '18 at 17:39
StubbornAtom
6,37831440
answered Dec 21 '18 at 12:10
José Carlos SantosJosé Carlos Santos
174k23133243
edited Dec 21 '18 at 17:39
StubbornAtom
6,37831440
edited Dec 21 '18 at 17:39
StubbornAtom
6,37831440
edited Dec 21 '18 at 17:39
StubbornAtom
6,37831440
6,37831440
answered Dec 21 '18 at 12:10
José Carlos SantosJosé Carlos Santos
174k23133243
answered Dec 21 '18 at 12:10
José Carlos SantosJosé Carlos Santos
174k23133243
answered Dec 21 '18 at 12:10
José Carlos SantosJosé Carlos Santos
174k23133243
174k23133243
$begingroup$
Thanks! Putting it like this I feel a bit stupid that I did not notice this myself...
$endgroup$
– User123456789
Dec 21 '18 at 12:17
1
$begingroup$
Believe me, I know that feeling. I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 21 '18 at 12:19
add a comment |
$begingroup$
Thanks! Putting it like this I feel a bit stupid that I did not notice this myself...
$endgroup$
– User123456789
Dec 21 '18 at 12:17
1
$begingroup$
Believe me, I know that feeling. I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 21 '18 at 12:19
$begingroup$
Thanks! Putting it like this I feel a bit stupid that I did not notice this myself...
$endgroup$
– User123456789
Dec 21 '18 at 12:17
$begingroup$
Thanks! Putting it like this I feel a bit stupid that I did not notice this myself...
$endgroup$
– User123456789
Dec 21 '18 at 12:17
1
1
$begingroup$
Believe me, I know that feeling. I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 21 '18 at 12:19
$begingroup$
Believe me, I know that feeling. I'm glad I could help.
$endgroup$
– José Carlos Santos
Dec 21 '18 at 12:19
add a comment |
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