Let $V$ be a finite inner product space and let $T:V \to V$ be a linear transformation. Prove that $T$ is diagonalizable if and only if the adjoint transformation $T^{*}$ is diagonalizable.

I got stuck with this problem. Since $(T^{*})^{*}=T$, it is sufficient to show the forward implication. Suppose $T$ is diagonalizable, then there exists a basis of eigenvectors $B=\{v_1,...,v_n\}$. I would like to show that $T^{*}$ also has a basis of eigenvectors.

By definition of $T^*$ We have $$\langle T(v_i),v_i \rangle=\langle v_i,T^*(v_i)\rangle$$

But $$\langle T(v_i),v_i \rangle=\alpha \langle v_i,v_i \rangle$$

I don't know what to do next, I would appreciate any hints. Thanks in advance.

  • 1
    $\begingroup$ Formulation in terms of matrix: if $T=BDB^{-1}$ then $T^*=(BDB^{-1})^*=(B^{-1})^*DB^*=(B^*)^{-1}DB^*$. (see here) $\endgroup$ – Surb Aug 26 '15 at 21:43
  • $\begingroup$ Thanks for your help. I think there is a tiny detail to correct: the field $K$ could be $\mathbb C$, so $T^*=(B^*)^{-1}\overline{D}B^*$, since $\overline{D}$ is a diagonal matrix, it follows $T^*$ is diagonalizable. $\endgroup$ – user16924 Aug 26 '15 at 22:00

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