Eigenvalues of special singular matrix












2












$begingroup$


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.










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$endgroup$

















    2












    $begingroup$


    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.










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      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.










      share|cite|improve this question









      $endgroup$




      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






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      asked Dec 21 '18 at 12:01









      User123456789User123456789

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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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












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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          2












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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




          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


















          • $begingroup$
            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


















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