Prove matrix $A$ is diagonalizable if and only if each eigenvalue has equal geometric and algebraic multiplicity.

Question

Prove: If $A$ is an $n\times n$ matrix with distinct eigenvalues $\lambda_1, \dots, \lambda_k$, then $A$ is diagonalizable if and only if $g_{\lambda_i}=a_{\lambda_i}$ for $1\le i \le k$.

I have seen and understood a proof for $\sum g_{\lambda_i}=n=\sum a_{\lambda_i}$ but nothing for the equality of each distinct eigenvalue.

Answer 1

Hint for one direction: If $A$ is diagonalizable as $A=PDP^{-1}$, then the characteristic polynomial of $A$ is the same as the characteristic polynomial of $D$ (why?). The latter is easy to compute and allows you to describe the algebraic multiplicities. The geometric multiplicities are also easy to describe, since you have all the eigenvectors (columns of $P$).

Hint for the other direction: if all the geometric and algebraic multiplicities match up, then you are able to construct an eigenbasis for $\mathbb{R}^n$ and directly build your diagonalization.

Answer 2

The converse is not valid in general for non-algebraically closed fields.

In fact, consider the rotation $T:\mathbb{R}^3\rightarrow\mathbb{R}^3$ around the $z$-axis. Its characteristic polynomial is

$$p_T=(X^2+1)(X-1).$$

So, 1 is the unique eigenvalue of $T$ and algebraic and geometric multiplicities of $1$ agree. However, $T$ is not diagonalizable since the eigenspace of $1$ has dimension $1<3$.

In general, you have to suppose that $p_T$ has a decomposing into linear factors. From now on, I think the previous post answers the question.