# Linear operator is compact if and only if its adjoint is compact

Let $$H$$ be a Hilbert space, and $$A:H\rightarrow H$$ a linear operator.

Prove that $$A$$ is compact if and only if $$A^*$$ is compact.

I saw the following proof in my book -

What I don't understand is, why the original theorem ($$A$$ is compact iff $$A^*$$ is compact) is equivalent to $$A$$ is compact iff $$A^*A$$ is compact?

Also - when it says that 'if $$A$$ is compact, then obviously $$A^*A$$ is compact', if I want to formally prove it:

If $$A$$ is compact, then for every $$\{x_n\}$$ that is bounded, we get that $$\{Ax_n\}$$ has a Cauchy subsequence,$$\{x_{n_k}\}$$.

Therefore, $$A*(f(x_{n_k})) = f(A(x_{n_k}))$$ for every $$f\in H^*$$. Since $$A,f$$ are bounded, there are both continuous (as it is equivalent to being bounded) then $$f(A(x_{n_k}))$$ is a Cauchy subsequence as well, meaning that $$A^*$$ is compact.

Is it correct? is it the way to prove it?

There is maybe a simpler way to show the second assertion. To say that an operator is compact is to say that the image of the unit ball is compact. Hence, if $$B$$ is the unit ball of $$H$$, then $$K:= A(B)$$ is compact and $$A^{*}A(B) = A^*(K)$$, which is compact since $$A$$ is continuous, so is $$A^*$$ and the image of a compact under a continuous map is compact.
Now, to explain the reasoning of the book. Suppose you have proven that "for any operator $$A$$, $$A$$ is compact iff $$A^*A$$ is compact". Call this theorem T1. Let us show that if $$A$$ is compact then $$A^*$$ is compact. Suppose $$A$$ is compact and set $$B=A^*$$. Since $$B^* = A$$ is compact, then $$B^*B$$ is compact (easy it is proven just like before). By Theorem $$T1$$, this shows that $$B$$ is compact, i.e. $$A^*$$ is compact.