4
$\begingroup$

Exercise: Suppose $V$ is finite dimensional and $T\in L(V)$ is such that $\|Tv\|\le \|v\|$ for every $v\in V$. Prove that $T-\sqrt{2}I$ is invertible.

Proof: We will prove the contrapositive. Suppose that $T-\sqrt 2I$ is not invertible. Thus, $\sqrt 2$ is an eigenvalue of $T$ and there exists a non-zero vector $v$ such that $Tv=\sqrt{2}v$. Taking the norm on both sides and dividing by $\|v\|$ we get $\frac {\|Tv\|} {\|v\|}=\sqrt 2$. Because $\sqrt 2>1$, the above equality implies that $\|Tv\|>\|v\|$ as if $0\le a < b$, then $\frac a b<1$.

Is this proof correct? Is there a better way to prove this?

Edit: For any future readers, the proof above could be better phrased and there is no need to bring up eigenvalues(although $\sqrt 2$ will still be an eigenvalue). I never explicitly state where I use the hypothesis that $V$ is finite dimensional. The following is what I would write as a proof now.

Proof: We will prove the contrapositive. Suppose that $T-\sqrt 2I$ is not invertible. Because $V$ is finite dimensional, injectivity implies invertibility. Thus, $T-\sqrt 2I$ is not injective and there exists a non-zero vector $v\in null(T-\sqrt 2I)$ such that $(T-\sqrt 2I)v=0\implies Tv=\sqrt 2v$. Taking the norm on both sides and dividing by $\|v\|$ we get $\frac {\|Tv\|} {\|v\|}=\sqrt 2$. Because $\sqrt 2>1$, the previous equality implies that $\|Tv\|>\|v\|$ as if $0\le a < b$, then $\frac a b < 1$.

$\endgroup$
16
  • $\begingroup$ What is $V$ and what is the norm on $V$? $\endgroup$ Commented Oct 17, 2022 at 8:03
  • 2
    $\begingroup$ Looks correct to me. Just a comment: this proof only works if $V$ is finite-dimensional. If $\text{dim }V=\infty$ and $T$ is a bounded linear operator on $V$, with $\|Tv\|\leq \|v\|$ for all $v$, then the same conclusion holds. $\endgroup$
    – Feng
    Commented Oct 17, 2022 at 8:03
  • 1
    $\begingroup$ @geetha290krm There was no specific norm specified. $V$ is taken to be an inner product space. $\endgroup$
    – Seeker
    Commented Oct 17, 2022 at 8:04
  • $\begingroup$ Your proof is fine as long as $V$ is finite dimensional. $\endgroup$ Commented Oct 17, 2022 at 8:14
  • 1
    $\begingroup$ @Seeker yeah. $T:V\to V$ is an linear operator on finite dimensional vector space. In linear algebra, we mostly study about finite dimensional vector space. Yes I meant to write $T-\lambda I$, not $T$. Which is “obvious” from context. Here is details of relation between bijectivity, injectivity and surjectivity, on a finite dimensional vector space. You have constructed lots of counter example, which is nice but those things you may have proved in past. $\endgroup$
    – user264745
    Commented Oct 18, 2022 at 11:07

0

You must log in to answer this question.

Browse other questions tagged .