Prove Eigen vectors of a Hermitian matrix form a non-singular matrix. 
I know that the Eigen vectors of a Hermitian matrix are orthogonal. 
But i can't move from there. Please help me.
 A: This is true for every matrix formed of eigenvectors.
Since eigenvectors corresponding to same or different eigenvalues are L.I.
In case of hermitian matrix we have $$A=PDP^{-1} $$
A: I assume that the $X_i$ are orthogonal with respect to the standard Hermitian inner product on $\Bbb C^n$, viz. $\langle V, W \rangle = \bar V \cdot W$; then:
Since, as the text of the problem states, the eigenvectors $X_i$ are orthogonal:
$\langle X_i, X_j \rangle = 0, \; i \ne j, \tag 1$
we can form the the matrix $C$ whose $k$-th column is $X_k$:
$C = \begin{bmatrix} X_1 & X_2 & \ldots & X_k & \ldots & X_n \end{bmatrix}, \tag 2$
and then form $C^\dagger$, the Hermitian adjoint of $E$, whose rows are the complex conjugates of the columns of $E$:
$C^\dagger = \begin{bmatrix} \bar X_1 \\ \bar X_2 \\ \vdots \\ \bar X_k \\ \vdots \\ \bar X_n \end{bmatrix}; \tag 3$
then it is easy to see that
$D = C^\dagger C = \begin{bmatrix} \langle X_i, X_j \rangle \end{bmatrix} = \begin{bmatrix} \delta_{ij} \Vert X_i \Vert^2 \end{bmatrix} \tag 4$
is a diagonal matrix whose $ii$ entry is $\Vert X_i \Vert^2 = \langle X_i, X_i \rangle \ne 0$.  Now $D$ is nonsingular, its inverse being
$D^{-1} = \begin{bmatrix} \delta_{ij} \Vert X_i \Vert^{-2} \end{bmatrix}; \tag 5$
it then follows that both $C$ and $C^\dagger$ are non-singular as well.
