About matrix products $A^{T}A$ and $ AA^{T} $

Question

I'm investigating the relationship between 2-norms and eigenvalues of $A^{T}A$ and $ AA^{T} $, in order to better understand the SVD decomposition.

How can I prove that $A^{T}A$ and $ AA^{T} $ are symmetric and positive definite?

Many thanks.

Answer 1

The matrix $A^TA$ is symmetric since $$\left( A^TA\right)^T=A^T\left(A^T\right)^T=A^TA$$ and if $A$ is invertible then $A^TA$ is definite positive since $$\langle A^TA x,x\rangle=\langle A x,Ax\rangle=\underbrace{||Ax||^2}_{\ge0}=0\iff Ax=0\iff x=0$$

The same proof is for the matrix $AA^T$.

Answer 2

A matrix, $M$, is symmetric if $M^T=M$. Let's concentrate on $A^TA$ for the moment, the case for $AA^T$ is very similar. So, all we need to do to test symmetry is to check whether $(A^TA)^T=A^TA$.

\begin{align} (A^TA)^T &=A^T(A^T)^T\\ &=A^TA\\ &\blacksquare \end{align}

A matrix $M$ is positive semidefinite if for any $x\in \Re^n$, $x^TMx \geq 0$. So, using the definition, let's see if the matrix is positive definite.

\begin{align} \mbox{Let } &x\in\Re^n\\ x^T(A^TA)x&=(Ax)^T(Ax)\\&=\|Ax\|_2^2\geq0 \end{align}

Answer 3

First, a matrix $B$ is symetric iff $B^T = B$.

Then, $B$ is positive definite iff $$ \forall x\in\Bbb R^d\ \ \ x^TBx \ge 0 \ \ \text{and} \ x^TBx =0 \implies x=0 $$