In an answer to this question I asked:Issue finding a unitary matrix which diagonalizes a Hermitian, it was said that without normalising my eigen vectors I wouldn't get the corresponding diagonal matrix containing the eigen values.
I am still very much learning this topic and can't see why my eigen vectors need to be of length 1, only then will putting them in a matrix and doing a similarity transform give me a diagonal matrix of the eigen values. So far in the course, this has never been pointed out or required for a successful diagonalisation of a matrix by similarity transform. I am guessing it is because in this case the matrix in question has complex elements.
If $A$ is Hermitian, and $\lambda_1, \lambda_2,\ldots, \lambda_n$ are distinct eigenvalues of $A$ then automatically all eigenvectors $v_i$ for $\lambda_i$ will be orthogonal (i.e. have inner product $0$) if they correspond to different eigenvalues. This is a standard theorem.
We can always multiply an eigenvector by a scalar and it will still be an eigenvector (for the same eigenvalue); eigenvectors for one eigenvalue form a linear subspace.
So we can make an orthogonal basis for each eigenspace and combine these to a full base (if we have $n$ eigenvectors and the matrix is $n \times n$) which is automatically orthogonal.
Then multiplying each vector $v$ by $\frac{1}{\|v\|}$ ensures the matrix $U$ of eigenvectors is orthonormal and $U^\dagger = U^{-1}$ which is easy in computations.