T a diagonalizable linear operator on V $\implies$ By the Complex Spectral Theorem, T is normal

diagonalizationgram-schmidtlinear algebrasolution-verification

Prove if the following statement is true.

Let V be a finite dimensional C-vector space with inner product and T a diagonalizable linear operator on V. Then there is a basis of eigenvectors of T for V. Applying the Gram-Schmidt orthogonalization process to this basis and then normalizing, an orthonormal basis of eigenvectors for V is obtained. By the Complex Spectral Theorem, T is normal.

Attempt:

T a diagonalizable linear operator on V if and only if T has a basis of n eigenvectors, in which case the diagonal entries are the eigenvalues for those eigenvectors. Let $$\beta$$ be that basis.

By Gram-Schmidt, $${\beta}'$$ is an orthonormal basis of V.

Since T is diagonalizable, $$[T]_{{\beta}'}$$ is diagonalizable.

There exists an orthonormal basis $${\beta}'$$ of V such that $$[T]_{{\beta}'}$$ is diagonalizable if and only if T is normal.

Therefore, the statement is true.

Am I wrong?

Sorry, I'm a bit confused. Is the question whether all diagonalizable linear operators $$T$$ on a finite dimensional complex vector space $$V$$ are normal? (i.e. whether you have given a proof of this?)
If so, unfortunately the answer is no. Such a linear operator $$T$$ is normal if and only if $$T$$ is unitarily diagonalizable (i.e. is diagonalized by conjugation with a unitary matrix), and this is just not the case in general.
To disprove the claim it suffices to just give an example, e.g. the matrix: $$\begin{bmatrix} 1&1\\ 0&2 \end{bmatrix}$$ is diagonalizable with eigenvalues $$1$$ and $$2$$ (check this), but is not normal.
You can try to apply your argument to this specific matrix to find out what's wrong. This issue is that $$[T]_{\beta'}$$ need not be a diagonal matrix, even though $$[T]_\beta$$ was. In my example matrix there is an eigenvector $$\begin{bmatrix} 1\\0 \end{bmatrix}$$ with eigenvalue $$1$$ and an eigenvector $$\begin{bmatrix} 1\\1 \end{bmatrix}$$ with eigenvalue $$2$$, but these vectors are not orthogonal. You might be relying on a "fact" that eigenvectors for distinct eigenvalues are orthogonal, but unfortunately this is not true in general (it holds for e.g. self-adjoint or normal matrices).
On the other hand, if we restrict $$T$$ to in addition be a self-adjoint linear operator then your argument does work, and you have given a valid proof that all self-adjoint linear operators are normal.