Why should I care that a matrix is diagonalizable?

Question

I was talking to someone about diagonalizability of matrices and they asked me why we even care. For one thing, we can easily raise the matrix to the $n$-th power.

If $A=PDP^{-1}$ then, $A^n = P D^n P^{-1}$ and the diagonal matrix $D$ is very easy to raise to the $n$-th power (just raise all the diagonals).

Ok, he said - so just a computational benefit.

Also, you can find the determinant by multiplying the numbers on the diagonal. Ok, again a computational benefit since you could find it through other means too.

I realized that I can't think of a fundamental benefit one gets from being able to diagonalize a matrix that extends beyond computational efficiency. For invertibility, there are many things you get that you just can't do otherwise. Is there anything similar one gets for diagonalizability that I'm not aware of?

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Note another similar question: A matrix is diagonalizable, so what? that asks for a solid perception of diagonalizability. I believe this one is slightly different since it focuses on any utility barring numerical efficiency.

Answer 1

You ask for a non-computational benefit of diagonalizing, but have you ever done anything with linear algebra that is not based on computations? Since you speak about matrices rather than linear maps, I wonder if you have ever used linear algebra outside of a computational setting.

Diagonalizability, when it occurs, gives you a good coordinate system to use for understanding a linear map (a set of axes built from eigenvectors). In particular, look up the spectral theorem! It is a very important result about diagonalizability of symmetric matrices.

There are many results in math called "the spectral theorem", for different kinds of linear maps (abstract versions of matrices), and they are very highly valued because being able to diagonalizable a matrix or a linear map is the mathematical analogue of decoupling a physical system (or a differential equation modeling a physical system) into parts that don't interact and can be analyzed separately.

Answer 2

Another reason: because many concepts in higher-level mathematics depend on it. Take, say, exponentiating a matrix, which is an important notion in higher-level geometry and Lie groups. A rough sketch of how to do this is as follows. Let $A = PDP^{-1}$ be an $n \times n$ diagonalisable matrix, where $D = \operatorname{diag}(d_1,d_2,\ldots,d_n).$ Then \begin{equation} \begin{split} e^A &= I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \ldots \\ &= I + PDP^{-1} + \frac{1}{2!} PD^2P^{-1} + \frac{1}{3!} PD^3P^{-1}+ \ldots \\ &= P \left[I + D + \frac{D^2}{2!} + \frac{D^3}{3!}+\ldots \right]P^{-1} \\ &= Pe^D P^{-1}, \end{split} \end{equation} where $e^D = \operatorname{diag}(e^{d_1}, e^{d_2},\ldots,e^{d_n}).$ Note that the end result is a simple expression involving the multiplication of three matrices, whereas the first line is an expression that is comparatively harder to evaluate; an infinite series involving terms of matrix powers.

It is possible to describe many other matrix functions in a similar way, including logarithm and trigonometric functions. Note that the diagonalisation is critical here; without it, we could not get past the first equality, and the whole concept of exponentiating a matrix wouldn't be worth much.

Answer 3

You could make the argument that it is also just for computational benefits, but diagonalization and related properties make solving simple systems of ODEs much easier.

Answer 4

It's not a big problem for most computations either, because the set of all diagonalizable matrices is dense in the set of all matrices. This means that if a particular way of simplifying a computation of a continuous expression only works for diagonalizable matrices, and you want to do the computation for a non-diagonalizable matrix, then you can consider a small perturbation around your matrix which then makes that perturbed matrix a diagonal one.

You then perform the computation for that case and then take the limit to zero perturbation in the result. This will always give the correct answer for continuous functions of the non-diagonalizable matrix.