# If $T$ is self-adjoint and $\alpha \in \mathbb{R}$, show that $\alpha \langle Tv,v \rangle \geq -|\alpha| \lVert Tv \rVert \lVert v \rVert$

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:

Proof:

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?

• Which specific step are you asking about? Which "we know that" do you disagree with? Sep 12 at 18:07
• 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
– GS2
Sep 12 at 18:09

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 $$\alpha \langle Tv,v\rangle \geq -|\alpha|\|Tv\|\|v\|.$$