Quesion(From Sheldon Axler's Linear Algebra Done Right, Section 7F, Exercise 11):
Let $V$ be a nonzero finite-dimensional inner product space over $\mathbb{C}$, and let $T \in \mathscr{L}(V)$ be a linear map. Prove that for every $\epsilon>0$， there exists a diagonalizable operator $S \in \mathscr{L}(V)$ such that $0<\|T-S\|<\epsilon， \forall v\in V$.
My idea: It is an easy problem if we don't need $S$ to be diagonalizable. Actually, we can use the single value decomposition(SVD) on $T$, say $Tv=s_{1}\langle v,e_{1}\rangle f_{1}+...+s_{m}\langle v,e_{m}\rangle f_{m}$, in which $s_{1}...s_{m}$are all the positive single values of $T$ (included as many times as the dimension of corresponding eigenspace of $T^{*}T$)and $f_{1}...f_{m}$ forms an orthonormal list of $T$. We can simply let $Sv=Tv+\frac{\epsilon}{2}\langle v,e_{1}\rangle f_{1}， \forall v\in V$ and the problem is solved.
However, this problem asks us to let $S$ be diagonalizable. How could this condition be satisfied? I have tried polar decomposition, but it seems doesn't work. If you have any idea or answer, please tell me, thanks! :)
Notation: $T^{*}$ stand for the adjoint of $T$.
