Proof for real symmertic, skew-symmetric matrices are always diagonalizable?

Question

Was reading a lot about certain kinds of matrices being always diagonalizable,

like, real symmetric, skew-symmetric, etc,

What would be the proof behind it ?

Answer 1

These are all normal matrices, i.e. matrices such that $AA^*=A^*A$. Normal matrices are diagonalizeable by the Spectral theorem, and in fact unitarily so.

The easiest way to prove this is to consider the Schur decomposition of $A$, which allows us to write $A=QUQ^*$ for some unitary matrix $Q$ and upper triangular matrix $U$. Since $$QUU^*Q^*=(QUQ^*)(QUQ^*)^*=(QUQ^*)^*(QUQ^*)=QU^*UQ^*$$ we have $UU^*=U^*U$. Comparing the first diagonal entries gives us $$\sum_{k=1}^n |u_{1k}|^2=|u_{1n}|^2$$ so $u_{1k}=0$ for $k\ne 1$. Comparing the next diagonal entries gives us $$\sum_{k=2}^n |u_{2k}|^2=|u_{12}|^2 + |u_{22}|^2 = |u_{22}|^2$$ so $u_{2k}=0$ for $k\ne 2$. Continuing in this manner, we see that $U$ must be diagonal.