# $A$ is positive definite iff there exists a nonsingular $C$ s.t. $C^*AC$ is positive definite

I am trying to prove the following statement for a matrix $A\in\mathbb{C}^{n\times n}$

$A$ is positive definite if and only if there exists a nonsingular matrix $C\in\mathbb{C}^{n\times n}$ such that $C^*AC$ is positive definite.

In the above, the fact that $A$ is Hermitian is implicitly assumed since all positive definite matrices are Hermitian. Also the asterisk denotes the Hermitian operator.

My attempt is the following. For the forward part, assume that $A$ is positive definite. Then choose $C$ as the identity matrix, $C=I_n$. Then, indeed $C^*AC=I^*AI=A$ is positive definite by assumption.

For the reverse, assume that $C^*AC$ is positive definite for some nonsingular $C\in\mathbb{C}^{n\times n}$. Let $\mathbf{x}\in\mathbb{C}^n$ and let $\mathbf{y}=C\mathbf{x}$. By assumption, $$0<\mathbf{x}^*C^*AC\mathbf{x}=\mathbf{y}^*A\mathbf{y}$$ i.e. $A$ is positive definite.

I am not sure whether my arguments are OK.

• I remember that $A$ is positive definite if $x^*Ax { \color{red}>} 0 \forall x$. – C.Ding Apr 10 '18 at 12:03

Let $y \ne 0$, then $\exists x \ne 0$ such that $y=Cx$ since $C$ is nonsingular.
Hence $$y^*Ay=(Cx)^*A(Cx)=x^*(C^*AC)x >0$$
• Can you please justify this argument more i.e. why do we need $C$ to be nonsingular? My intuition is that is this was not the case, then we could not guarantee one-to-one correspondence between $\mathbf{x}$ and $\mathbf{y}$. – mgus Apr 10 '18 at 16:20
• In the first place if $C$ is singular, then we can find $x \ne 0$ such that $Cx=0$, hence $C^*AC$ is not positive definite. what we have to show is that $\forall y \ne 0, y^*Ay>0$, and this holds because $\forall y \ne 0, \exists x \ne 0$ such that $y=Cx$. – Siong Thye Goh Apr 10 '18 at 16:31