Let $M$ be an invertible $n\times n$ matrix and $N$ a symmetric $n \times n$ matrix.
Prove that $M^TNM$ is positive definite if and only if $N$ is positive definite.
My thoughts on this: Suppose that $N$ is symmetric positive definite.
$N$ is symmetric, therefore there is an orthogonal matrix $M$ (orthogonality implies invertibility) such that $M^{-1}NM = M^TNM = D$. Where $D$ is the diagonal matrix of $N$.
$N$ is also positive definite, and this implies that all its eigenvalues are positive, i.e., $D$ is a positive definite matrix.
Therefore, $M^TNM$ is positive definite.
However, I don't know if my proof is "formal" enough. Moreover, I haven't proven the other way around (it's an iff statement). Could someone help me on this?
Thanks!
Best Answer
A symmetric matrix $A\in \mathbb{R}^{n\times n}$ is positive definite iff $x^TAx>0$ for arbitrary vector $x\in \mathbb{R}^n$.
Sufficiency: Consider $\forall x\in\mathbb{R}^n$. $x^TM^TNMx=(Mx)^TN(Mx)=x_1^TNx_1$. Since $N$ is pd, $x_1^TNx_1>0$. Hence $M^TNM$ is pd.
Necessity: Consider $\forall x\in\mathbb{R}^n$. $x^TNx=x^T(M^{-1})^TM^TNMM^{-1}x=x_1^TM^TNMx_1$. Since $M^TNM$ is pd, $x_1^TM^TNMx_1>0$. Hence $N$ is pd.