Prove that $v\in V$ such that $\| v\|=1$ and $\| T(v)\|_W = \max\limits_{\Vert u \Vert_V = 1} \Vert T(u) \Vert_W$ is an eigenvector of $T^*T$

Question

Let $V,W$ be finite-dimensional complex inner product spaces and $T:V \to W$ a linear map.

Let $v\in V$ such that $\| v\|=1$ and $\| T(v)\|_W = \max\limits_{\Vert u \Vert_V = 1} \Vert T(u) \Vert_W$.

Prove that $v$ is an eigenvector of $T^*T$

I try this direction, but I am really not sure $\| T(v)\|^2_W=\langle T(v),T(v) \rangle = \langle v,T^*T(v) \rangle$

Please any help or suggestion. Thanks.

Answer 1

For any $u$ with $\|u\|_V=1$, $$\vert\langle T^*Tv, u \rangle \vert=\vert \langle Tv, Tu \rangle \vert\le \lVert Tv\rVert_W \lVert Tu\rVert_W\le \lVert Tv\rVert_W \lVert Tv\rVert_W$$ as by hypothesis

$$\lVert T(u) \rVert_W \le \| T(v)\|_W = \max\limits_{\Vert u \Vert_V = 1} \Vert T(u) \Vert_W.$$

As all inequalities turn into equalities for $u = v$,

$$\sup\limits_{\Vert u \Vert_V = 1} \lvert\langle T^*Tv, u \rangle \rvert$$ is attained when $u=v$.

We can decompose $u = \lambda_u T^*Tv + z$ where $z$ is orthogonal to $T^*Tv$. Then if $\lVert T^*Tv \rVert_W \neq 0$

$$ \lvert\langle T^*Tv, u \rangle \rvert = \lvert \lambda_u \rvert \lVert T^*Tv \rVert_W$$ is maximum when $\lvert \lambda_u \rvert$ is maximum. As $$1=\lVert u \rVert^2 = \lvert \lambda_u \rvert^2 \lVert T^*Tv \rVert^2+ \lVert z \rVert^2$$ we need to have $z=0$ and $u = \lambda T^*Tv$. Which means that $v$ is an eigenvector of $T^*T$.

And if $\lVert T^*Tv \rVert_W =0$, then $T$ is the zero linear map and the desired result holds trivially.

Answer 2

Hint

$U = T^*T: V \to V$ is a self-adjoint operator. Hence is diagonalizable in an orthonormal basis of eigenvectors. Moreover, the eigenvalues are nonnegative real numbers.

Now, you can order the eigenvalues $\lambda_1 \lt \lambda_2 \lt \dots \lt \lambda_p$ of $U$ and decompose $$v = v_1 + v_2 + \dots + v_p$$ over the associated eigenspaces. Finally, prove that $v_1 = v_2 = \dots = v_{p-1}=0$ using the condition

$$\| T(v)\|_W = \max\limits_{\Vert u \Vert_V = 1} \Vert T(u) \Vert_W$$ and

$$\| T(v)\|^2_W=\langle T(v),T(v) \rangle = \langle v,T^*T(v) \rangle$$

With this, you're done.