I was hoping to get some help on a review problem from my text book. The problem reads:

Let $A$ be $m\times{n}$ matrix. Prove that

a) $A^*A$ is self-adjoint.

b) All eigenvalues of $A^*A$ are non-negative.

c) $A^*A + I$ is invertible.

For part a), it seems to follow very clearly that $(A^* A)^* =(A)^* (A^*)^* =A^*A$. So this first part is clear.

With parts b) and c) I am unsure how to begin. For b), would one want to take the inner product of $(A^*{A}x,x)$ and then say some $Ax=qx$ where $q$ is an eigenvalue and then get $(Ax,Ax)=(qx,qx)= |q|^2\|x\|^2$? So the eigenvalue is positive? That is what I am thinking but I am unsure if it shows what I am wanting to show/ does not have a ton of holes. With part c) I am just lost so would appreciate any advice possible on that one. Thank you!

Upon more thought, I have a new idea for b). So we have some A* Ax =qx. So (Ax,Ax)=(A* Ax,x)=q(x,x). So q=((Ax,Ax)/(x,x))=||Ax||/||x||, which is greater than or equal to zero, so non-negative. would that work?


2 Answers 2


For part (b), you are on the right track but you want to adopt a different approach. The question is about the eigenvalues of $A^*A$, not those of $A$. So suppose $x$ is an eigenvector of $A^*A$, i.e. $A^*Ax = qx$ for some scalar $q$. Now you can use your approach (use the inner product) to deduce things about $q$.

For part (c), there are a few things you could do. The easiest thing to do would be to show that the equation $(I+A^*A)x = 0$ has only the trivial solution. Use linearity to expand things out, and rearrange. How are $x$ and $A^*A$ related? Can you use the previous parts to say anything?

(There is also a high-powered proof of (c) using the spectral theorem - you can show that $I+A^*A$ has strictly positive eigenvalues, and therefore positive determinant. But that's akin to killing a mosquito with a sledgehammer.)

  • $\begingroup$ So for b) would it be something like saying x is an eigenvalue of AA, so AAx=qx. So if we took the inner product (A*Ax,x)=(qx,x)=|q||x|| or something like that or am I taking it in the wrong direction? $\endgroup$ Nov 15, 2016 at 21:54
  • $\begingroup$ actually that would not work without (A* Ax,A* Ax) so would I take then inner product of the two of them to get that q is pos? $\endgroup$ Nov 15, 2016 at 21:56
  • $\begingroup$ Your first approach is correct, but you're making a mistake: $(qx,x)\neq|q|\|x\|^2$. You should figure out the correct expression. Then you can apply the definition of $A^*$ to $(A^*Ax,x)$... $\endgroup$ Nov 16, 2016 at 3:04
  • $\begingroup$ would it be the square root of q? so then since we have the square root of q, q has to be positive? $\endgroup$ Nov 16, 2016 at 5:02
  • $\begingroup$ Don't guess, go back to the definition and properties of the inner product. If $a$ is a scalar and $x,y$ are vectors, what is $(ax,y)$? $\endgroup$ Nov 16, 2016 at 5:15

For part (b), if

$A^\dagger A \vec v = \mu \vec v, \; \vec v \ne 0, \tag 1$


$\mu \langle \vec v, \vec v \rangle = \langle \vec v, \mu \vec v \rangle = \langle \vec v, A^\dagger A \vec v \rangle = \langle A\vec v, A \vec v \rangle \ge 0; \tag 2$


$\langle \vec v, \vec v \rangle > 0, \tag 3$

(2) yields

$\mu = \dfrac{\langle A\vec v, A \vec v \rangle}{\langle \vec v, \vec v \rangle} \ge 0. \tag 4$

For (c), if $A^\dagger A +I$ is not invertible, then

$\exists \vec v \ne 0, \; A^\dagger A \vec v + \vec v = (A^\dagger A + I) \vec v = 0; \tag 5$


$A^\dagger A \vec v = -\vec v, \tag 6$


$\langle \vec v, A^\dagger A \vec v \rangle = -\langle \vec v, \vec v \rangle; \tag 7$


$\langle \vec v, A^\dagger A, \vec v \rangle = \langle A\vec v, A\vec v \rangle \ge 0, \tag 8$


$-\langle \vec v, \vec v \rangle < 0; \tag 9$

combining (7), (8), and (9) yields

$0 \le \langle \vec v, A^\dagger A \vec v \rangle = -\langle \vec v, \vec v \rangle < 0, \tag{10}$


$0 < 0; \tag{11}$

this contradictory statement shows that there can be no such $\vec v \ne 0$; hence

$\vec v = 0, \tag{12}$

and thus $A^\dagger A + I$ is invertible.


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