# 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 $T^*T$

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.

• Are you working in finite-dimensional spaces? Feb 2 at 9:45
• Do you mean $\|T(v)\|_W$ instead of $\|T(v)\|_w$? Feb 2 at 9:49
• I am working in finite-dimensional spaces , and I mean $\|T(v)\|_w$ , since $T: V \to W$
– algo
Feb 2 at 9:51
• @erez $\Vert T(v) \Vert_w$ doesn't make sense as the name of the space is $W$, not $w$. Feb 2 at 9:53
• Sorry , it suppose to be $W$
– algo
Feb 2 at 10:01

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.

• The point is that you're using the equality case of Cauchy-Schwarz. This is fine to get as @daw mentioned the equality $Tu = Tv$. But that is not sufficient to justify your final claim Hence $T^*Tv$ is a scalar multiple of $v$. Or at least an additional argument is required. Feb 2 at 10:40
• I'm using a basic fact that $|(u,v)|$ as a function of $u$ on the unit sphere achieves maximum iff $u$ is a scalar multiple of $v$. This is clearly the case for real finite dimensional inner product space because the angle between the two vectors has to be $0$ or $\pi$. And similar arguments can be used to prove the complex case. Feb 2 at 10:45
• I got it! That maybe deserves an argument in your answer. What is nice is that this would also work for infinite-dimensional space. Feb 2 at 10:50
• Why it is correct : $|Tv\| \|Tu\|\le \|Tv\|\|Tv\|$ ?
– algo
Feb 2 at 11:53
• @Justauser I allowed myself to edit your answer with additional elements. Feb 2 at 13:35

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.