To prove a matrix non singular

Question

I know I have to prove determinant of a matrix as non zero. But how to proceed.

Please guide.

Thanks a lot in advance.

Answer 1

A Hermitian matrix $A$ admits the following decomposition (diagonalisation): $$A V = V \Lambda\tag{*}$$ Where $\Lambda=\textrm{diag}(\lambda_1,...,\lambda_n)$ and $V=[X_1,...X_n]$. We also know that the columns of the matrix $V$ are linearly independent (proof below), $i.e.$ $\det{(V)}\neq 0$.

Because $A$ is Hermitian $V^TV=I$ where $I$ is the identity matrix., therefore: $$\det(V)=1/\det{(V^T)}$$ From $(*)$ $$\det{(A)} = \det{(V\Lambda V^T)}=\det{(\Lambda)}=\prod_i^n{\lambda_i}$$ Therefore $A$ would be nonsingular if $\lambda_i\neq 0$

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To prove that $V$ is nonsingular you can show that in the case $A$ is Hermitian the equations $$AV=V\Lambda \qquad UA=\Lambda U$$ show that $U=V^T$, and therefore the matrix $V$ is orthogonal:

Answer 2

Let $i\ne j$. We have: $\newcommand\inner[2]{\langle #1, #2 \rangle}$

$$\lambda_i\inner{X_i}{X_j} = \inner{\lambda_i X_i}{X_j} = \inner{AX_i}{X_j} = \inner{X_i}{AX_j} = \inner{X_i}{\lambda_j X_j} = \overline{\lambda_j}\inner{X_i}{X_j}$$

Assume $\inner{X_i}{X_j} \ne 0$. By cancelling $\inner{X_i}{X_j}$ we get $\lambda_i = \overline{\lambda_j}$, and since the eigenvalues of a hermitian matrix are real, $\lambda_i = \lambda_j$. But, this is a contradiction with all the eigenvalues being distinct.

$\inner{X_i}{X_j} = 0$ follows, so the set $\{X_1, \ldots, X_n\}$ is orthogonal and thus linearly independent. This implies that $C$ is nonsingular.

Answer 3

The columns are linearly independent, since they are eigenvectors that correspond to distinct eigenvalues. Therefore, the matrix is non-singular.