I'm studying Sheldon Axler's "Liner Algebra done right" book, but I'm having some trouble understanding the proof of Lemma 7.11.

Lemma (7.11): Suppose that $T \in \mathcal{L}(V)$ is self-adjoint. If $\alpha, \beta \in \mathbb{R}$ are such that $\alpha^2 < 4 \beta$ then $T^2+ \alpha T + \beta I$ is invertible.

The book basically shows that if $v$ is a nonzero vector of $V$ then $\langle T^2 v + \alpha T v + \beta v , v \rangle >0$ and hence $T^2 v + \alpha T v + \beta v \neq 0$ so $\text{Nul}(T^2 + \alpha T + \beta I) = \{ 0 \}$ and therefore $T^2 + \alpha T + \beta I$ is invertible:


Let $v$ be a nonzero vector in $V$. Then:

$\langle T^2 v + \alpha T v + \beta v, v\rangle = \langle T^2v, v \rangle + \alpha \langle Tv,v \rangle + \beta \langle v,v \rangle$

$ = \langle Tv, Tv \rangle + \alpha \langle Tv,v \rangle + \beta \lVert v \rVert ^2$ (Since $T$ is self-adjoint)

$= \lVert Tv \rVert^2 + \alpha \langle Tv,v \rangle + \beta \lVert v \rVert ^2$

And then the book says that using Cauchy-Schwarz inequality we get:

$\lVert Tv \rVert^2 + \alpha \langle Tv,v \rangle + \beta \lVert v \rVert ^2 \geq \lVert Tv \rVert^2 - |\alpha| \lVert Tv \rVert \lVert v \rVert + \beta \lVert v \rVert ^2$

But I can't understand why the last inequality holds (Specifically, I do not understand why the inequality $\alpha \langle Tv,v \rangle \geq -|\alpha| \lVert Tv \rVert \lVert v \rVert$ is satisfied.). Using Cauchy-Schwarz inequality we know that:

$\langle Tv, v \rangle \leq \lVert Tv \rVert \lVert v \rVert$ and hence $- \langle Tv, v \rangle \geq -\lVert Tv \rVert \lVert v \rVert$.

We also know that $\alpha \geq -|\alpha|$. If one assumes that $\langle Tv,v \rangle \geq 0$ then we can mutply $\alpha \geq -|\alpha|$ to get: $\alpha \langle Tv,v \rangle \geq -|\alpha| \langle Tv,v \rangle \geq -|\alpha| \lVert Tv \rVert \lVert v \rVert$, but the problem does not give information about the sing of $\langle Tv,v \rangle$ (in fact It could be a negative number), so why does $\alpha \langle Tv,v \rangle \geq -|\alpha| \lVert Tv \rVert \lVert v \rVert$ holds? Is it a mistake in the book?

  • $\begingroup$ Which specific step are you asking about? Which "we know that" do you disagree with? $\endgroup$
    – user7530
    Sep 12 at 18:07
  • $\begingroup$ I don't understand how does $\alpha \langle Tv,v \rangle \geq -|\alpha| \lVert Tv \rVert \lVert v \rVert$ can be obtained using Cauchy Schwarz inequality $\endgroup$
    – GS2
    Sep 12 at 18:09

1 Answer 1


I don't understand how does $\alpha \langle Tv,v\rangle \geq −|\alpha|\| Tv\|\|v\|$ can be obtained using Cauchy Schwarz inequality

Vanilla Cauchy-Schwarz gives you $|\langle Tv,v\rangle| \leq \|Tv\|\|v\|.$ Multiplying both sides by $-|\alpha|$ gives $$-|\alpha||\langle Tv,v\rangle| \geq -|\alpha|\|Tv\|\|v\|.$$ Now $$\alpha \langle Tv,v\rangle \geq -|\alpha \langle Tv,v\rangle| = -|\alpha||\langle Tv,v\rangle|$$ so $$\alpha \langle Tv,v\rangle \geq -|\alpha|\|Tv\|\|v\|.$$


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