A matrix is diagonalizable, so what?

Question

I mean, you can say it's similar to a diagonal matrix, it has $n$ independent eigenvectors, etc., but what's the big deal of having diagonalizability? Can I solidly perceive the differences between two linear transformation one of which is diagonalizable and the other is not, either by visualization or figurative description?

For example, invertibility can be perceived. Because non-invertible transformation must compress the space in one or more certain direction to $0$. Like crashing a space flat.

Answer 1

Up to change in basis, there are only 2 things a matrix can do.

1. It can act like a scaling operator where it takes certain key vectors (eigenvectors) and scales them, or
2. it can act as a shift operator where it takes a first vector, sends it to a second vector, the second vector to a third vector, and so forth, then sends the last vector in a group to zero.

It may be that for some collection of vectors it does scaling whereas for others it does shifting, or it can also do linear combinations of these actions (block scaling and shifting simultaneously). For example, the matrix $$ P \begin{bmatrix} 4 & & & & \\ & 3 & 1 & & \\ & & 3 & 1 &\\ & & & 3 & \\ & & & & 2 \end{bmatrix} P^{-1} = P\left( \begin{bmatrix} 4 & & & & \\ & 3 & & & \\ & & 3 & &\\ & & & 3 & \\ & & & & 2 \end{bmatrix} + \begin{bmatrix} 0& & & & \\ & 0& 1 & & \\ & & 0& 1 &\\ & & & 0& \\ & & & &0 \end{bmatrix}\right)P^{-1} $$ acts as the combination of a scaling operator on all the columns of $P$

• $p_1 \rightarrow 4 p_1$, $p_2 \rightarrow 3 p_2$, ..., $p_5 \rightarrow 2 p_5$,

  plus a shifting operator on the 2nd, 3rd and 4th columns of $P$:

• $p_4 \rightarrow p_3 \rightarrow p_2 \rightarrow 0$.

This idea is the main content behind the Jordan normal form.

Being diagonalizable means that it does not do any of the shifting, and only does scaling.

For a more thorough explanation, see this excellent blog post by Terry Tao: http://terrytao.wordpress.com/2007/10/12/the-jordan-normal-form-and-the-euclidean-algorithm/

Answer 2

The behavior of linear dynamical systems, both continuous and discrete, can be expressed in terms of the eigenvalues of the relevant matrix, and the expression (and especially the long-term behavior) has some added complications if the matrix is not diagonalizable.

Answer 3

If an $n\times n$ real matrix $M$ is diagonalizeable, then it corresponds to stretching $\mathbb R^n$ along $n$ linearly independent lines some factor $\lambda_n$. Otherwise this is not the case.

Answer 4

I'll try an answer in a different (equivalent) direction: what happens when the matrix is not diagonalizable?

First of all, this must mean that some of the matrix's eigenvalues occur more than once. Otherwise the matrix really can't do anything else than simply stretching its eigenvectors by $\lambda_n$. So what if two eigenvalues are equal? Let's write down the most non-trivial non-diagonalizable example there is $$\pmatrix{0 & 1 \cr 0 & 0}.$$ This (as you have correctly observed) crushes the space. But the important point is that it doesn't crush it to zero! Instead it only crushes it to some subspace. In general, if you have a nilpotent matrix (all eigenvalues vanish) there are many subspaces (of varying dimensions) to pick from and so many different ways to crush the space. The zero operator sends everything to zero immediately while it will take a nontrivial nilpotent matrix some (finite) time in a sense that for $T$ nilpotent, $T^n = 0$ for some $n$. In general, every nilpotent matrix is similar to a matrix that looks something like this $$\pmatrix{0 & 0 & 1 & 0 \cr 0 & 0 &1 &1 \cr 0 & 0 & 0 & 1 \cr 0 & 0 & 0 &0}$$ with some ones above the diagonal. The precise position of those ones (if there are any) determines which subspace is being crush to which and so on. You are very much encouraged to play with such matrices in $\mathbb R^3$ (where there are not too many possibilties how a nilpotent matrix can look like, but still enough to show what happens).

Having said all this, it would be a sin now not to mention Jordan decomposition. When studying a matrix $A$ you first find its eigenvalues and corresponding eigenspaces. So pick such an eigenspace corresponding to the eigenvalue $\lambda$. Then $A - \lambda$, restricted to this eigenspace, is a nilpotent operator! If this nilpotent operator is zero, then the original matrix $A$ just stretches this eigenspace by $\lambda$. But in general it can perform a lot of nontrivial shuffling (corresponding to the nilpotent part).