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This is from Linear Algebra Done Right 4th edition 7C Exercise 7.

The question goes like this:

V is a nonzero finite-dimensional inner product space.

Suppose is an invertible positive operator on V and is a positive operator on V. Prove that + is invertible.

My current approach:

Because S is invertible and positive, I know there exists an orthonormal basis of V with respect to which S has an diagonal matrix with only positive number on the diagonal. And S^(-1) is also positive.

I know that S + T is also positive, and to show that it is invertible I can try to find an orthonormal basis of V with respect to which S + T has an diagonal matrix with only positive number on the diagonal. But I have trouble constructing one. What am I missing? Or is there other simpler approach?

Tim
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2 Answers2

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(Thanks @Tzimmo for providing the hint.)

For all $v \in V$ and $v\not= 0$, we have

$$ \langle (S+T)v , v \rangle = \langle Sv , v \rangle + \langle Tv , v \rangle$$

Since $S$ is positive, if $\langle Sv , v \rangle = 0,$ $Sv = 0$. Since $S$ is invertible, $ Sv\not=0 $ and thus $\langle Sv , v \rangle > 0$.

Since $T$ is positive, $\langle Tv , v \rangle \geq 0$

So $ \langle (S+T)v , v \rangle > 0$.

This tells us $ null(S+T) = \{0\} $, i.e. $S+T$ is injective. As $S+T$ is an operator, the fact that it is injective implies it is invertible.

Tim
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Expanding on Tim's previous answer. Once we get that $\operatorname{null}(S+T) = \{0\}$, we then get that

$$\operatorname{range}T^* = (\operatorname{null}T)^\perp = V$$

but since $T*=T$ (since it is positive), we get that $T$ is surjective.

Hyperbolic Cake
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