To show $I+ A$ is non singular

Question

$A$ is a positive operator on Hilbert space $H$, I have to show the title of this question.

Since $A $ is positive so all eigenvalues are $\ge 0$, so eigenvalues of $I+A$ are $\ge 1$, so $\det(I+A) \ne 0$, hence non singular.

Is my proof correct? Another way to show that the operator is a bijection on $H$, I have done that it is injection but could any one show that this is onto also?

Thanks for helping.

score 2 · Accepted Answer · answered Jun 14 '15 at 13:13

2

Your proof using determinant is only valid in finite-dimensional spaces or for special classes of $A$, see Fredholm determinant.

Since $A$ is positive, $$ ( (I+A)x,(I+A)x) = \|x\|^2 + \|Ax\|^2 + 2(Ax,x) \ge \|x\|^2 + \|Ax\|^2. $$ Hence $I+A$ is injective. Moreover, $(I+A)^{-1}: Im(I+A)\to H$ is bounded (set $x:=(I+A)^{-1}y$, $y\in Im(I+A)$).

Next, we show that the range/image of $I+A$ is closed. Take a sequence $y_n=(I+A)x_n$, $x_n\in H$, converging to $y$. Then by the above observation, $(I+A)^{-1}y_n$ is a Cauchy sequence, hence $(x_n)$ is converging to $x$. This implies $ y = (I+A)x$ by continuity of $I+A$.

Last step is to show surjectivity. Take $z \in (Im(I+A))^\perp$. Then it holds $$ 0 = (z, (I+A)z) \ge \|z\|^2, $$ hence $z=0$. Since $Im(I+A)$ is closed this proves $Im(I+A)=H$, and $I+A$ is continuously invertible.

This proof mimics the proof usually given to prove the Lax-Milgram theorem.

answered Jun 14 '15 at 13:13

daw

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I don't understand your proof sorry – Myshkin Jun 14 '15 at 17:15
What exactly? Which points are not clear? – daw Jun 14 '15 at 17:41
How you showed injection and surjection – Myshkin Jun 15 '15 at 01:55
Injectivity: if $(I+A)x=0$ then the first inequality implies $|x|=0$. Surjectivity: first I prove that the range is closed. Second, that the orthogonal complement of the range is ${0}$. Exactly which arguments are unclear? – daw Jun 15 '15 at 06:19

score 1 · Answer 2 · answered Jun 14 '15 at 08:09

First of all, determinants in a Hilbert space are a whole can of worms you don't want. There's no reason why $I+A$ even has a determinant. If $A=I$, for example. There is no determinant for $2I$.

An operator is invertible if and only if 0 is not in its spectrum. So that's all you really need to show.

Your approach will work, but you need to say more about why the spectrum of $I+A$ had the lower bound you describe. The sum of two operators need not behave that well in general. Use the positivity of $A$ to make it work.

To show $I+ A$ is non singular

2 Answers2