# 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?

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Every vector is an eigenvector of $A$ with common eigenvalue $\lambda$; in particular, the unit vectors are. So... – David Mitra Jun 16 '13 at 23:45
@Kaster, No, I meant diagonal. A 3x3 diagonalisable matrix could have 3 corresponding eigenspaces each of dimension 1. – user46080 Jun 16 '13 at 23:48
Here's a hint: What are the three dimensional subspaces of $\mathbb{R}^3$? (Or $\mathbb{F}^3$ for any non-trivial field $\mathbb{F}$?) – Tom Oldfield Jun 16 '13 at 23:55
@David Then let $P = \begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$. Hence $A = PDP^{-1} = IDI = D$. Hence A is diagonal? – user46080 Jun 17 '13 at 0:00
Yes, you can do that. But I meant to imply that $Ae_i=\lambda e_i$ for each $i=1,2,3$. But $Ae_i$ is the $i$th column of $A$. The result follows directly. – David Mitra Jun 17 '13 at 0:02

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?

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This seems like an unnecessarily convoluted way to approach this problem. – Tom Oldfield Jun 16 '13 at 23:53
+1: I think this is a reasonable approach? – copper.hat Jun 17 '13 at 0:46
Dear @TomOldfield : That assessment seems rather overblown considering the solution is two sentences containing two simple steps, one of which is commonly used when analyzing such problems. I'm sorry you felt strongly enough about it to memorialize in a comment :( – rschwieb Jun 17 '13 at 2:21
@copper.hat Thanks for the vote of confidence! – rschwieb Jun 17 '13 at 2:22
thanks for the response! – user46080 Jun 17 '13 at 4:14

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.

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This doesn't seem any easier to prove than the original statement, to me at least... – Tom Oldfield Jun 16 '13 at 23:51
@TomOldfield: It was just showing the OP how to formalize the answer. – copper.hat Jun 16 '13 at 23:54
I suppose. But the OP's attempt talked about parameters, which is completely different in spirit to this approach, so I think that, in a sense, this isn't formalising their answer as much as providing a different one. This is all fine, but the "different" answer looks more convoluted than other possible "different" answers. No offence meant... – Tom Oldfield Jun 17 '13 at 0:04
@TomOldfield: None taken! I suppose I view 'free parameters' as synonymous with dimensions... – copper.hat Jun 17 '13 at 0:08
Makes sense, thanks! – user46080 Jun 17 '13 at 0:12

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.

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The eigenvalues are 3, 2 and 1, so we have three 1-dimensional eigenspaces, not one 3-dimensional eigenspace, which was my question. I can't why your answer is relevant, could you explain please? – user46080 Jun 17 '13 at 0:20
@user46080 OK, I've messed up with eigenspace and the space spanned by eigenvectors. And I've updated my answer. If you feel it somehow helpful, you can cancel the downvote. – ziyuang Jun 17 '13 at 0:29
Wasn't my downvote dood! – user46080 Jun 17 '13 at 0:43
@user46080 Well, OK, calm... – ziyuang Jun 17 '13 at 0:43