Is every set of independent eigenvectors of an orthogonally diagonalizable matrix orthogonal?

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Let $f : V rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $mathbbC$ of dimension $n$.
Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1ldots,u_n$, corresponding to the eigenvalues $lambda_1,ldots,lambda_n$.
Now suppose $v_1,ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $lambda_1,ldots,lambda_n$ respectively.



Question: Are these "new" eigenvectors also orthogonal?



If yes, please provide a proof. If not, a counterexample.



To avoid the trivial case, assume that at $lambda_1,ldots,lambda_n$ are NOT pairwise distinct.






linear-algebra eigenvalues-eigenvectors







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  • This reads like a verbatim homework assignment. Did you try anything at all?
    – Umberto P.
    Sep 6 at 11:55














up vote
1
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Let $f : V rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $mathbbC$ of dimension $n$.
Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1ldots,u_n$, corresponding to the eigenvalues $lambda_1,ldots,lambda_n$.
Now suppose $v_1,ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $lambda_1,ldots,lambda_n$ respectively.



Question: Are these "new" eigenvectors also orthogonal?



If yes, please provide a proof. If not, a counterexample.



To avoid the trivial case, assume that at $lambda_1,ldots,lambda_n$ are NOT pairwise distinct.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question





















  • This reads like a verbatim homework assignment. Did you try anything at all?
    – Umberto P.
    Sep 6 at 11:55












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $f : V rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $mathbbC$ of dimension $n$.
Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1ldots,u_n$, corresponding to the eigenvalues $lambda_1,ldots,lambda_n$.
Now suppose $v_1,ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $lambda_1,ldots,lambda_n$ respectively.



Question: Are these "new" eigenvectors also orthogonal?



If yes, please provide a proof. If not, a counterexample.



To avoid the trivial case, assume that at $lambda_1,ldots,lambda_n$ are NOT pairwise distinct.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question













Let $f : V rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $mathbbC$ of dimension $n$.
Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1ldots,u_n$, corresponding to the eigenvalues $lambda_1,ldots,lambda_n$.
Now suppose $v_1,ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $lambda_1,ldots,lambda_n$ respectively.



Question: Are these "new" eigenvectors also orthogonal?



If yes, please provide a proof. If not, a counterexample.



To avoid the trivial case, assume that at $lambda_1,ldots,lambda_n$ are NOT pairwise distinct.






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






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asked Sep 6 at 11:48









EpsilonDelta

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  • This reads like a verbatim homework assignment. Did you try anything at all?
    – Umberto P.
    Sep 6 at 11:55
















  • This reads like a verbatim homework assignment. Did you try anything at all?
    – Umberto P.
    Sep 6 at 11:55















This reads like a verbatim homework assignment. Did you try anything at all?
– Umberto P.
Sep 6 at 11:55




This reads like a verbatim homework assignment. Did you try anything at all?
– Umberto P.
Sep 6 at 11:55










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Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.






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    1 Answer
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    up vote
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    accepted










    Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.






    share|cite|improve this answer
























      up vote
      3
      down vote



      accepted










      Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.






      share|cite|improve this answer






















        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.






        share|cite|improve this answer












        Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 6 at 11:51









        John Hughes

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