Prove that the linear function $f$ in $V$ is diagonalizable iff there exists a base of $V$ composed of eigenvectors of $f$

Question

Quote from my textbook:

Definition: A linear function $f : V \rightarrow V$ is diagonalizable if there is a base $\mathscr{B}$ of $V$ such that the representative matrix $M_\mathscr{B}(f)$ is a diagonal matrix.

From the definition immediately follows that a linear function $f: V \rightarrow V$ is diagonalizable if and only if there exists a base $\mathscr{B}$ of $V$ composed of eigenvectors of $f$.

I can't come up with a proof for it, though. Help?

Answer 1

Suppose there exists a basis $\mathscr{B}:=\{v_1,\ldots,v_n\}$ of $V$ consisting of eigenvectors of $f$. Then $f(v_i)=\lambda_i v_i$ for each eigenvector, so the matrix $M_{\mathscr{B}}(f)$ is a diagonal matrix whose $i$th diagonal entry is $\lambda_i$.

Conversely, suppose $f$ is diagonalizable. Then there exists a basis $\mathscr{B}:=\{v_1,\ldots,v_n\}$ such that the matrix $M_{\mathscr{B}}(f)$ is diagonal, say with diagonal entries $d_1,\ldots,d_n$. Then $f(v_i) = M_{\mathscr{B}}(f) e_i = d_i e_i$ so $v_i$ is an eigenvector of $f$.

Answer 2

Hint: Given a linear map $T:U \rightarrow V$ between (same dimension) vector spaces, how can you write $T$ as a matrix? Remember that you take the basis of $U$ and write the image (under $T$) of each one as a combination of the basis elements of $V$. What does it mean if this proccess gives a diagonal matrix? What's the definition of an eigenvector?