# Show that operator is normal and determine its Singular Value Decomposition

Consider the operator $$Af(x):=\int\limits_{-\pi}^{\pi}\sin(x-y)f(y)\, dy, x\in [-\pi,\pi], f\in L_2(-\pi,\pi).$$ Show that the operator $A\in\mathcal{L}(L_2(-\pi,\pi))$ is normal. Determine the Singular Value Decomposition (SVD) of A.

In order to check if A is normal, I determined the adjoint operator with the result that $$A^* f(x)=\int\limits_{-\pi}^{\pi}\overline{\sin(x-y)} f(y)\, dy.$$

Then I calculated $AA^*$ and $A^*A$. Here are my results: $$AA^* f(x)=\int\limits_{-\pi}^{\pi}\sin(x-y) \int\limits_{-\pi}^{\pi}\overline{\sin(y-z)} f(z)\, dz\, dy$$ $$A^*Af(x)=\int\limits_{-\pi}^{\pi}\overline{\sin(x-y)}\int\limits_{-\pi}^{\pi}\sin(y-z)f(z)\, dz\, dy$$ And this is identical because $\sin(x)=\overline{\sin(x)}$.

Could you please write me in a comment if it is okay until now?

Thanks a lot.

Edit:

Concerning the SVD:

Is it right, that I have to determine the eigenvalues (resp. eigenfunctions) of $AA^*$? Does one need the convolution theorem of the Fouriertransformation? Explicitly: To my opinion it is $$AA^*f(x)=(\sin\star A^*f)(x)$$ and therefore $$AA^* f(x)=\lambda f(x)\Leftrightarrow \mathcal{F}(\sin\star A^*f)=(2\pi)^{1/2}\mathcal{F}(\sin)\cdot\mathcal{F}(A^*f)=\lambda\mathcal{F}(f),$$ i.e. $$(2\pi)^{1/2}\mathcal{F}(\sin)\cdot\mathcal{F}(A^*f)=\lambda\mathcal{F}(f).$$ Can one use that equation to determine now $\lambda$ resp. $f$?

Greetings

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Here is a related problem for normal matrices. – Mhenni Benghorbal Feb 12 '13 at 11:03

Yes, Fourier transform is the way to go.

Recall the spectral theorem: a bounded operator on a Hilbert space is normal if and only it is unitarily equivalent to a multiplication operator. What you have here is a special case, with the implementing unitary being the Fourier transform $\mathcal{F}$.

As you noted, the convolution theorem says $\mathcal{F}$ maps convolution to multiplication. The Pontryagin dual in this case, $\mathbb{Z}$, is discrete. This means $A$ is diagonalizable, in particular normal. The $\mathcal{F}(h)$ for any $h \in L^1(\mathbb{T})$ is a sequence vanishing at infinity. So $A$ is compact and SVD makes sense.

In Fourier domain, your answer is very simple: $A$ is projection onto the basis element $\sin x$ (up to a scaling factor, possibly). So the eigenvalues of $A^*A$ is same as $A$: $1$ for the basis element projected on and $0$'s for every other element (again, you might need to adjust for a scaling factor that makes $\mathcal{F}$ unitary).

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Note that, if $$K \equiv \int_{a}^{b}k(x,t) dt$$ and

$$H \equiv \int_{a}^{b}h(x,t) dt,$$ then

$$KH \equiv \int_{a}^{b}k(x,z)h(z,t) dz.$$

Now, apply this result to your operators $A$ and $A^{*}.$

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Why is KH that way? – math12 Feb 12 '13 at 15:48