Suppose that a $3\times 3$ matrix $M$ has an eigenspace of dimension $3$. Prove that $M$ is a diagonal matrix. How would I go about this? I realise that having dimension 3 means that the solution to $(A-\lambda I)\mathbf b = \mathbf 0$ has 3 free parameters, which would in turn mean that $(A-\lambda I)$ is the zero matrix, so $A = \begin{bmatrix}\lambda & 0 & 0\\0 & \lambda & 0\\0 & 0 & \lambda\end{bmatrix}$, but the first part of that argument is a bit too heuristic. How would I prove it formally?
 A: Hint: Imagine you have three linearly independent (column) eigenvectors for $\lambda$, and put them into the columns of a matrix $B$, which is necessarily nonsingular. Then $AB=B(\lambda I_3)=(\lambda I_3)B$. 
Can you see how to conclude this, remembering that $B$ has an inverse?

In general, no matter what the eigenvalues are, you can use this observation to find a matrix $B$ such that $AB=BD$ where $D$ has the eigenvalues of $A$ on the diagonal, and is zero elsewhere. But unless all the eigenvalues are the same, you can't always conclude that $BD$ and $DB$ are the same. They are quite often different. Can you see why?
A: If any matrix $B$ satisfies $B v_k = 0$ for a basis $v_k$, then $B=0$.
In this case, the dimension of the space is $3$ and the dimension of the eigenspace is $3$, hence we have $(A-\lambda I)v_k = 0$ on a basis $v_k$, hence
$A = \lambda I$ as required.
A: Note that $A$ can be diagonalized $\Longleftrightarrow$ $A$'s algebraic multiplicity $=$ $A$'s geometric multiplicity. And according to the definition of similarity, $A$ itself is also a diagonal matrix.
The intuition behind the theorem is that there is no need to resort to Jordan chain because there are "enough" eigenvectors. And since there is no Jordan chain, the matrix can be diagonalized.
