Show that $T^*T$ is positive semi definite

Question

I am struggeling with this one:
Let $X = \mathbb{R}^n$, $Y = \mathbb{R}^m$. We equipe $X$ with the scalare product $\langle x_1,x_2 \rangle_X = x_2^T \cdot M_x \cdot x_1 $, where $M_x\in \mathbb{R}^{n \times n}$ spd. Analogously, let $\langle y_1,y_2 \rangle_Y = y_2^T \cdot M_y \cdot y_1 $, where $M_y\in \mathbb{R}^{m \times m}$ spd.

Define the operator:

$$T:X \rightarrow Y, Tx = Ax, \quad \text{ with } A\in \mathbb{R}^{n\times m}.$$

One can show that $$T^*y = A_*y = M_x^{-1} \cdot A^T \cdot M_y\cdot y$$.

Now I wanted to show that $T^*T$ is positive semi-definite, i.e.

$$\begin{align*} x^T \cdot T^*T\cdot x \geq 0 \text{ for all } x\in X. \end{align*}$$

This is were I struggle. During my first attempt, I totally forgot that $M_x$ and $M_y$ are not the identity matrix. So I computed:

$$\begin{align*} x^T \cdot T^*T \cdot x = \langle T^*T x, x\rangle_X = \langle Tx, Tx\rangle_Y = \lVert Tx\rVert_Y \geq 0. \end{align*}$$

Now I really hoped that this would work for the above setting in a similar way:

$$\begin{equation*} x^T\cdot T^*\cdot T \cdot x = x^T \cdot A_* \cdot A \cdot x = x^T \cdot M_x^{-1} \cdot A^T \cdot M_y \cdot x \end{equation*}$$

but I don't know how to continue. Can somehone help me out?

Thanks in advance:)

Answer 1

You just need to finish the computation.

The computation is \begin{align} \langle T^*Tx,x\rangle_X&=x^T\, M_x\, T^*T\, x_1 =x^T\, M_x M_x ^{-1} A^T M_y A\, x\\[0.3cm] &=x^T(A^TM_yA)x=(Ax)^TM_y Ax\geq0, \end{align} since $M_y$ is positive definite.

Answer 2

It is true in more general setting.

$V, W$ be two inner product spaces over the same field $F$.

Let $T\in \mathcal{L}(V, W) $ , then $T^{\star}\in\mathcal{L}(W, V) $.

$T^{\star}T\in\mathcal{L}(V)$ is self adjoint operator as $(T^{\star}T) ^{\star}=T^{\star}(T^{\star})^{\star}=T^{\star}T$

$\begin{align}\langle T^{\star}Tv, v\rangle _V&=\langle Tv,Tv\rangle_W\\&=\|Tv\|_W^2\\&\ge 0\end{align}$

Hence $T^{\star}T$ is positive semi definite operator.