1
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ You can't assume that $M$ is orthogonal, and you don't need to. Are you familiar with the characterization of positive-definiteness in terms of inner products? $\endgroup$
    – Qiaochu Yuan
    Jun 16, 2011 at 13:47
  • $\begingroup$ Could you enlighten me on that? $\endgroup$
    – John
    Jun 16, 2011 at 13:50
  • $\begingroup$ Let $\langle \cdot, \cdot \rangle$ be the standard inner product on $\mathbb{R}^n$. There are two things you need to know that make this question trivial: first, $N$ is positive-definite if and only if $\langle v, Nv \rangle > 0$ for all nonzero $v$. Second, the transpose is characterized by the identity $\langle v, Mw \rangle = \langle M^T v, w \rangle$. $\endgroup$
    – Qiaochu Yuan
    Jun 16, 2011 at 13:55
  • $\begingroup$ I'm not familiar with these definitions, but thanks a lot for your suggestions. How do we proceed from the definitions? What about the symmetricity and invertibility? $\endgroup$
    – John
    Jun 16, 2011 at 14:10
  • $\begingroup$ Simply observe that $\langle v, M^T NM v \rangle = \langle Mv , NMv \rangle$. $\endgroup$
    – Qiaochu Yuan
    Jun 16, 2011 at 14:14

1 Answer 1

Reset to default
5
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .