In some textbooks it is mentioned that, if a matrix is Positive Semidefinite, then all its diagonal entries are nonnegative and all the principal submatrices of PSD obtained by removing any number of rows and corresponding columns of it, will also leads to PSD always.
Now my question is if the parent PSD matrix is diagonalizable, then what can be said about the principal submatrices. Will they be always diagonalizable?
A matrix is positive semi-definite iff all its eigenvalues are real and non-negative, but an $N\times N $ matrix $A$ is diagonalizable iff the eigenvectors form a basis for $\mathbb{R}^N$. So you know that $\det(A_{1:k,1:k}) \ge 0$ because the eigenvalues are non-negative, and $x'A_{1:k,1:k}x \ge0$. But you can have multiple (in particular, zero) eigenvalues, reducing the dimension of the space spanned by the associated eigenvectors, so that $A_{1:k,1:k}$ may not diagonalizable.