I know that the spectral theorem tells us that in the case of a real inner product space, an operator is self adjoint if and only if there is an orthonormal basis with only eigenvectors of that operator and that in the case of a complex inner product space, an operator is normal if and only if there is an orthonormal basis with only eigenvectors.
However, I am unable to see what is so important about this. I know that this means I can diagonalize the matrix but I don't know what is so good about diagonalizing a matrix. Also, why do we have to have an orthonormal basis? Why can it not be a standard basis?
Thank you!
Thanks to the spectral theorem you can easily compute the powers of a given diagonalizable matrix $A = P^{-1}D P$ (as @Gerry Myerson said), thanks to the relation $$A^k = P^{-1}D^k P.$$ But you can also exploit the diagonalization to evaluate more complex matrix functions, e.g. the matrix exponential. Indeed you have $$e^A = \sum_{k=0}^\infty \frac{A^k}{k!} = P^{-1}\sum_{k=0}^\infty \frac{D^k}{k!} P = P^{-1}e^D P,$$ and for diagonal matrices $D = \text{diag}(d_1,\ldots,d_n)$ we have the simple relation $$e^D = \text{diag}(e^{d_1},\ldots,e^{d_n}).$$ This holds in general for any matrix function defined as a power series, indeed if you have $f(x)=\sum_{k=0}^\infty c_k x^k$, then $$f(A) = P^{-1} f(D) P,$$ where $f(D) = \text{diag}(f(d_1),\ldots,f(d_n)).$
[EDIT: notice that the matrix exponential is extremely important, because thanks to this tool you can e.g. solve linear differential systems of equations $\dot{\mathbf{x}}(t) = A(t) \mathbf{x}(t) + \mathbf{b}(t)$.]