# Exercise $5$, Section $6.A$ - Linear Algebra Done Right

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 $$\frac a b < 1$$.

• What is $V$ and what is the norm on $V$? Commented Oct 17, 2022 at 8:03
• 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.
– Feng
Commented Oct 17, 2022 at 8:03
• @geetha290krm There was no specific norm specified. $V$ is taken to be an inner product space. Commented Oct 17, 2022 at 8:04
• Your proof is fine as long as $V$ is finite dimensional. Commented Oct 17, 2022 at 8:14
• @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. Commented Oct 18, 2022 at 11:07