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Let $\mathcal{M}_n(\mathbb{R})$ denote the vector space of matrices of size $n\times n$ on $\mathbb{R.}$.

For any $n\in\mathbb{N}$, one can find a basis of $\mathcal{M}_n(\mathbb{R})$ where all the matrices are non diagonalizable. However, can one find a basis of $\mathcal{M}_n(\mathbb{R})$ composed of diagonalizable matrices only ?

(Of course, those matrices don't have to be diagonalizable in the same basis)

For $n=2$, such a basis can be found; however for greater $n$s, I haven't found a general answer.

Let $E_{i,j}=((\delta_{i,j}))$ be an elementary matrix in $\mathcal{M}_n(\mathbb{R})$.

Some thoughts : one could easily gather $n(n+1)/$2 symmetric matrices to start a basis, namely the $E_{i,j}+E_{j,i}$ for $i\neq j$ and the $E_{i,j}$.

Had we been working in $\mathcal{M}_n(\mathbb{C})$, we could have added $n(n-1)/2$ antisymmetric $E_{i,j}-E_{j,i},i\neq j$ to get our basis, but that doesn't hold on $\mathcal{M}_n(\mathbb{R})$.

Hippalectryon
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    Well, at least over $\mathbb C$, diagonalizable matrices are dense. It seems likely to me that in any normed vector space, if you have a basis, it's possible to perturb each basis vector by some small amount and still end up with a basis (just visualize it in $\mathbb R^3$). So take a basis and then adjust each matrix in it slightly so as to be diagonalizable. – Jack M Nov 10 '15 at 20:37

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Here's one way to do this. I'll just give the idea, and if it makes sense then you can probably work out the details.

One particular class of diagonalizable matrices are those that have distinct eigenvalues. In general it's difficult to tell what the eigenvalues of a matrix are a matrix are, but for some classes of matrices it's very simple. For instance, the eigenvalues of an upper or lower triangular matrix are just the diagonal entries.

So, take a basis consisting of upper triangular and lower triangular matrices. Choose them so that each has distinct entries on the diagonal.

user134824
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  • How so ? I thought about this, but I can only see $n(n+1)/2$ symmetric matrices to put in the basis (Call $E_{i,j}$ the matrix with a $1$ on row $i$ column $j$ and $0$ elsewhere, then you can chose the $E_{i,j}+E_{j,i},i\neq j$ and the $E_{i,i}$) – Hippalectryon Nov 10 '15 at 20:19
  • What I said before was wrong (whoops!), but I have a new idea. – user134824 Nov 10 '15 at 20:32
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This would be a corollary of the (true) claim that diagonalizable $n \times n$ matrices form a dense subset of the space of $n \times n$ matrices, say, in the natural norm, $\|A\|^2 = \sum_{i,j} |a_{i,j}|^2$. This is because in that case you could take any orthonormal basis $\{e_{ij}\}_{1 \leq i,j\leq n}$, and perturb it slightly into a basis $\{e'_{ij}\}_{1 \leq i,j\leq n}$ which consists of diagonalizable matrices. Indeed, some thinking should show that $\sum_{i,j} \|e_{ij} - e'_{ij}\| < 1$ is a sufficient condition for the $e'_{ij}$ to form a basis, given that the $e_{ij}$ form an orthonormal basis.

Bartosz Malman
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