If A is invertible and orthogonally diagonalizable, is $A^{-1}$ orthogonally diagonalizable as well?












0












$begingroup$


I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?



What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?






linear-algebra matrices diagonalization







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you need a proof of the claim in the title?
    $endgroup$
    – user258700
    May 1 '16 at 23:25












  • $begingroup$
    Yes @AhmedHussein
    $endgroup$
    – Jet
    May 1 '16 at 23:25






  • 1




    $begingroup$
    If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
    $endgroup$
    – Greg Martin
    May 2 '16 at 5:02
















0












$begingroup$


I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?



What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?






linear-algebra matrices diagonalization







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you need a proof of the claim in the title?
    $endgroup$
    – user258700
    May 1 '16 at 23:25












  • $begingroup$
    Yes @AhmedHussein
    $endgroup$
    – Jet
    May 1 '16 at 23:25






  • 1




    $begingroup$
    If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
    $endgroup$
    – Greg Martin
    May 2 '16 at 5:02














0












0








0





$begingroup$


I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?



What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?






linear-algebra matrices diagonalization







share|cite|improve this question











$endgroup$




I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?



What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?






linear-algebra matrices diagonalization




linear-algebra matrices diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 2 '16 at 4:16









Martin Sleziak

44.8k9118272




44.8k9118272










asked May 1 '16 at 23:17









JetJet

12




12












  • $begingroup$
    Do you need a proof of the claim in the title?
    $endgroup$
    – user258700
    May 1 '16 at 23:25












  • $begingroup$
    Yes @AhmedHussein
    $endgroup$
    – Jet
    May 1 '16 at 23:25






  • 1




    $begingroup$
    If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
    $endgroup$
    – Greg Martin
    May 2 '16 at 5:02


















  • $begingroup$
    Do you need a proof of the claim in the title?
    $endgroup$
    – user258700
    May 1 '16 at 23:25












  • $begingroup$
    Yes @AhmedHussein
    $endgroup$
    – Jet
    May 1 '16 at 23:25






  • 1




    $begingroup$
    If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
    $endgroup$
    – Greg Martin
    May 2 '16 at 5:02
















$begingroup$
Do you need a proof of the claim in the title?
$endgroup$
– user258700
May 1 '16 at 23:25






$begingroup$
Do you need a proof of the claim in the title?
$endgroup$
– user258700
May 1 '16 at 23:25














$begingroup$
Yes @AhmedHussein
$endgroup$
– Jet
May 1 '16 at 23:25




$begingroup$
Yes @AhmedHussein
$endgroup$
– Jet
May 1 '16 at 23:25




1




1




$begingroup$
If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
$endgroup$
– Greg Martin
May 2 '16 at 5:02




$begingroup$
If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
$endgroup$
– Greg Martin
May 2 '16 at 5:02










1 Answer
1






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1












$begingroup$

Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
$$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
$$begin{pmatrix}
2 & 0 \
0 & 3
end{pmatrix} ^{-1} =
begin{pmatrix}
frac{1}{2} & 0 \
0 & frac{1}{3}
end{pmatrix}$$
Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.



Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.






share|cite|improve this answer









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    $begingroup$

    Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
    $$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
    $$begin{pmatrix}
    2 & 0 \
    0 & 3
    end{pmatrix} ^{-1} =
    begin{pmatrix}
    frac{1}{2} & 0 \
    0 & frac{1}{3}
    end{pmatrix}$$
    Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.



    Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
      $$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
      $$begin{pmatrix}
      2 & 0 \
      0 & 3
      end{pmatrix} ^{-1} =
      begin{pmatrix}
      frac{1}{2} & 0 \
      0 & frac{1}{3}
      end{pmatrix}$$
      Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.



      Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
        $$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
        $$begin{pmatrix}
        2 & 0 \
        0 & 3
        end{pmatrix} ^{-1} =
        begin{pmatrix}
        frac{1}{2} & 0 \
        0 & frac{1}{3}
        end{pmatrix}$$
        Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.



        Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.






        share|cite|improve this answer









        $endgroup$



        Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
        $$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
        $$begin{pmatrix}
        2 & 0 \
        0 & 3
        end{pmatrix} ^{-1} =
        begin{pmatrix}
        frac{1}{2} & 0 \
        0 & frac{1}{3}
        end{pmatrix}$$
        Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.



        Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 2 '16 at 5:28









        Jimmy XiaoJimmy Xiao

        141111




        141111






























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