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.

  • 2
    $\begingroup$ Are you working in finite-dimensional spaces? $\endgroup$ Feb 2 at 9:45
  • $\begingroup$ Do you mean $\|T(v)\|_W$ instead of $\|T(v)\|_w$? $\endgroup$ Feb 2 at 9:49
  • $\begingroup$ I am working in finite-dimensional spaces , and I mean $\|T(v)\|_w$ , since $T: V \to W$ $\endgroup$
    – algo
    Feb 2 at 9:51
  • 1
    $\begingroup$ @erez $\Vert T(v) \Vert_w$ doesn't make sense as the name of the space is $W$, not $w$. $\endgroup$ Feb 2 at 9:53
  • $\begingroup$ Sorry , it suppose to be $W$ $\endgroup$
    – algo
    Feb 2 at 10:01

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Feb 2 at 10:40
  • $\begingroup$ 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. $\endgroup$ Feb 2 at 10:45
  • $\begingroup$ I got it! That maybe deserves an argument in your answer. What is nice is that this would also work for infinite-dimensional space. $\endgroup$ Feb 2 at 10:50
  • $\begingroup$ Why it is correct : $|Tv\| \|Tu\|\le \|Tv\|\|Tv\|$ ? $\endgroup$
    – algo
    Feb 2 at 11:53
  • $\begingroup$ @Justauser I allowed myself to edit your answer with additional elements. $\endgroup$ Feb 2 at 13:35


$U = T^*T: V \to V$ 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.


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