Is any square matrix over $\Bbb{C}$ diagonalisable?

Question

I know the answer to the question is false (or so my professor said) but I am not sure why. I read http://www.imsc.res.in/~kapil/papers/matrix/ but did not quite understand it enough to answer my question. In the conclusion he said

"If we choose our matrix randomly (in a uniform distribution) from within a bounded region, then it will turn out to be diagonalisable over the complex numbers with probability 1 (in other words always)."

So it would seem to me that it should be true, unless one of his conditions listed (it being chosen in a uniform distribution from within the bounded region) is the reason that is becomes true. In which case I assume one of those conditions not always being met is the reason that the statement is false in general. Any help would be appreciated.

Answer 1

If an $n\times n$ matrix has $n$ distinct eigenvalues, then it is diagonalizable. In $\mathbb R^n$, the set of $n$-tuples $(x_1,\ldots,x_n)$ with repeated entries has measure zero (think of the cases $n=2$ and $n=3$ to get an intuition).

Answer 2

In probability theory, an event occurring with probability 1 is very different from "this will always happen". Instead it means "this will almost always happen". In this case, it comes down to the statement that diagonalisable matrices are dense in $\mathbb{C}^{n \times n}$.

To build a non-diagonalisable matrix, you could simply make a matrix that is in Jordan form (only entries on the diagonal and ones on the superdiagonal) with at least one 1 on the superdiagonal. In fact, matrices that are non-diagonalisable are precisely those that have such a Jordan form.

Answer 3

There is a very simple condition for diagonalisability over any field $K$:

A square matrix of dimension $n$ always satisfies a polynomial equation of degree at most $n$ — this because of Hamilton-Cayley's theorem: any square matrix $A$ is a root of its characteristic polynomial $\;\chi_A(x)=\det(A-xI)$ ($I$ is the unit matrix in dimension $n$).

So there exists a polynomial of minimal degree $m_A(x)\in K[x]$ such that $A$ is a root of $m_A$ (in the ring of $n\times n$ matrices). The criterion is the following:

$A$ is diagonalisable over $K$ if and only if its minimal polynomial splits into a product of distinct linear factors.

Now while any polynomial over $\mathbf C$ splits into a product of linear factors, it is not true that its factors will always be distinct.

Answer 4

An $n\times n$ matrix represents a linear operator on some vector space $V$. That is, it is the matrix representation with respect to some basis for $V$. A linear operator is diagonalizable if and only if there exists a basis for the space $V$ consisting of eigenvectors of the operator.

Consider the following matrix representation of the operator $T$: $$\mathcal{M}(T)=\begin{pmatrix} \lambda_1 & 1 & 0 & 0 \\ 0 & \lambda_1 & 1& 0 \\ 0 & 0 & \lambda_1 & 0\\ 0 & 0 & 0 & \lambda_2 \end{pmatrix} .$$ This is not a diagonalizable matrix/operator, because there exists no basis of eigenvectors of $T$ for $V$. There does exist a basis of generalized eigenvectors for $T$, which allows us an upper-triangular representation representation. In particular, the above is the Jordan form of the operator, but if you don't know about that yet, don't worry.

So, we are guaranteed an upper-triangular matrix representation for a given operator. That is, a matrix over a complex space is guaranteed to be upper-triangularizable. This is useful because the determinant is easy to compute, and the eigenvalues are displayed with multiplicity on the diagonal.

The reason why we care so much about this basis consisting of eigenvectors is that under $T$, and eigenvector $v_j$ satisfies: $$ Tv_j=\lambda_jv_j$$ where $\lambda_j$ is the corresponding eigenvalue. Because we know that the matrix records how the basis of the space is transformed, if we have a basis of eigenvectors $v_1,v_2,...,v_n$ where $n=dim V$, we know that $Tv_1,Tv_2,...,Tv_n=\lambda_1v_1,\lambda_2v_,...,\lambda_nv_n$ which yields a matrix of the form: $$ \begin{pmatrix} \lambda_1 & 0 & 0 & 0 \\ 0 & \lambda_2 & 0& 0 \\ 0 & 0 & \ddots & 0\\ 0 & 0 & 0 & \lambda_n \end{pmatrix} .$$ This is a diagonal representation of the operator.