What is the point of using the formula $P^{-1}AP =D$ for diagonalization?

Question

I recently learnt what diagonalization is, but I don’t get the point of using the formula $D = P^{-1}AP$ for diagonalization. If we know the eigenvalues after finding them can’t we just put them into a diagonal matrix straight away? What use is there in computing P and $P^{-1}$?

Answer 1

If I want to calculate powers of $A$ then $A^n=(PDP^{-1})^n=PD^nP^{-1}$ (why?) and since powers of $D$ are easy to calculate this means $A^n$ is easy to calculate as long as we know $P$. So that's why we calculate $P$.

Answer 2

Extending @CyclotomicField's answer, let $f(x)$ be a function with a Taylor series (about $0$), $$f(x)=\sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!},$$ valid for $x$ in some radius of convergence. We wish to define $f(A)$, where $A$ is a matrix; a simple way to do this would be to use the Taylor series for $f$. To efficiently compute $A^n$, we diagonalize $A=P^{-1}DP$ so that $$A^n=P^{-1}D^nP.$$ Thus we can efficiently compute $$f(A)=P^{-1}\left(\sum_{n=0}^\infty f^{(n)}(0)\frac{D^n}{n!}\right)P=P^{-1}f(D)P,$$ where $f(D)$ is simply applying $f$ to the diagonal elements (note that we require all of the elements to lie in the in radius of convergence). This has many applications, although a neat one is solve systems of differential equations: $$\frac{d\mathbf{x}}{dt}=A\mathbf{x}\implies \mathbf{x}(t)=e^{At}\mathbf{x}(0),$$ where $e^{At}$ can be computed using the method above.

Answer 3

Suppose you have a matrix $A$ and a vector $v$, and suppose $D$ is a diagonal version of $A$. If you compute $A v$, you will most likely not get the same result as if you compute $D v$.

Usually you compute the diagonal version of a matrix because it is useful for you in a specific expression. That is why you must in that case substitute $A$ by $P^{-1} D P$ to get an equivalent expression.