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Let $V=\Bbb F^{m\times n}$ $T: V\to V$ , by $T(B)=P^{-1}BP$ , for any $B$ in $V$ , where $P$ is an invertible matrix. prove that if $A$ is an eigenvector of $T$, with eigenvalue $\lambda$ and $A$ is a diagonalizable matrix, then $\lambda=1$.

I know that $A$ and $\lambda A$ are similar, then they have the same eigenvalues. then if $a$ is an eigenvalue of $A$, $(\lambda*a)$ is an eigenvalue of $(\lambda A)$, I'm not sure that I can claim then $(\lambda*a)=a$, because we don't know they have same eigenvectors.

Thank you in advance.

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Suppose that $A \neq 0$ is diagonalizable, then there exists an invertible (unitary) matrix $U$ such that $A = U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}$.

Then $\lambda A = T(A) = P^{-1}U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}P$.

We get $\lambda U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1} = P^{-1}U\mathrm{diag(\lambda_{1}, ..., \lambda_{n})}U^{-1}P$ (*)

On taking the trace of the (*) and using the fact $\mathrm{Tr(AB)} = \mathrm{Tr(BA)}$, we should be able to force the condition we need on $\lambda$. Better yet, when are two diagonal matrices of the same size similar?

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  • $\begingroup$ Good, you could also play with the same operator by considering other restrictions. One of those is that if that $P$ is also diagonalizable, then you show that T must be diagonalizable too and its eigenvalues could be found in terms of those of P. $\endgroup$ – akech Aug 22 '14 at 18:58
  • $\begingroup$ from the equation * you mentioned , I get: λ1+λ2+...+λn=0 or λ=1. if we also use det. then we have λ1λ2...λn=0 or λ=1. how we can conclude that λ=1? Thanks $\endgroup$ – Shirly Aug 22 '14 at 20:41
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    $\begingroup$ Is it clear from the equations in the hint that $\mathrm{diag}(\lambda \lambda_{1},..., \lambda \lambda_{n})$ is similar to $\mathrm{diag}(\lambda_{1},..., \lambda_{n})$? Now we have two diagonal matrices, which are similar! $\endgroup$ – akech Aug 22 '14 at 23:14
  • $\begingroup$ Yes, from the similarity we conclude that they have same eigenvalues: {λλ1,...,λλn}={λ1,...,λn}, therefore |λ|=1, and λ can be also -1, no? $\endgroup$ – Shirly Aug 23 '14 at 5:31
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I'm sorry, I think I found a contradiction. Let

$$A= \left(\begin{array}{ccc}1 & 0 & 0\\0 &-1& 1\\0& 0 &0 \end{array}\right)$$

A is diagonalizable.

Then if

$$P= \left(\begin{array}{ccc}0 & 1 & -1\\1 &0& 1\\0& 0 &1 \end{array}\right)$$

We get $P=P^{-1}$, and $P^{-1}AP=-A$, therefore $T(A)=-A$, and $\lambda=-1$.

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