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 $\frac a b < 1$.