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I'm trying to solve this problem:

Let $\Omega$ a bounded open of $\mathbb{R}$, consider the Hilbert real spaces $X = L^2(\Omega)$ and $Y = \mathbb{R}^{2\times 2}$, with the inner products: $$\langle u,v\rangle_X\ =\ \int_{\Omega}uv\quad\forall\ u,v\in X, \quad\quad \langle A,B\rangle_Y\ =\ \mathrm{tr}(A^TB)\quad \forall\ A,B\in Y,$$ and define the operator $\mathcal{A}:X\rightarrow Y$ by $$\mathcal{A}(u)\ :=\ \left(\begin{array}{cc} \int_{\Omega}u & \int_{\Omega}xu\\ \int_{\Omega}xu & \int_{\Omega}x^2u \end{array}\right),\quad \forall\ u\in X.$$ Show that $\mathcal{A}\in\mathcal{L}(X,Y)$ and compute explicitly the operator $\mathcal{A}^*$ and the spaces $\mathrm{Ker}(\mathcal{A})$, $\mathrm{Im}(\mathcal{A})$, $\mathrm{Ker}(\mathcal{A}^*)$ and $\mathrm{Im}(\mathcal{A}^*)$.

I have already proved that $\mathcal{A}\in\mathcal{L}(X,Y)$ when I consider functionals $F_j(u) = \langle u, x^j\rangle$, $j=0,1,2$ in $X^{\prime}$ and then define $$\mathcal{A}(u)\ :=\ \left(\begin{array}{cc} F_0(u) & F_1(u)\\ F_1(u) & F_2(u) \end{array}\right),\quad \forall\ u\in X.$$ So, I have that $$\mathcal{A}^*(P)\ =\ P_{11} + (P_{12} + P_{21})x + P_{22}x^2,\quad \forall\ P\in Y.$$ Now, I need help in the last part, I have proved that $$\mathrm{Ker}(\mathcal{A})\ =\ ^{\circ}\{F_0,F_1,F_2\}\ =\ \{1,x,x^2\}^{\perp}\quad \text{ and }\quad \mathrm{Ker}(\mathcal{A}^*)\ =\ \{P\in Y : P^T = -P\},$$ but I don't know how compute $\mathrm{Im}(\mathcal{A})$ and $\mathrm{Im}(\mathcal{A}^*)$.

Please somebody help me.

Thanks in advance

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Closed range theorem maybe? –  Shuhao Cao Jun 2 '13 at 3:50
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up vote 2 down vote accepted

For the domain of $\newcommand{\A}{\mathcal{A}}\A$ is all $X$, closed range theorem says: $$ \mathrm{Im}(\A) = \mathrm{Ker}(\A^*)^{\perp} = \{A\in Y: \mathrm{tr}(A^TB)=0,\;\forall B\in \mathrm{Ker}(\A^*)\}, $$ which are all symmetric $2\times 2$ matrices and $$ \mathrm{Im}(\A^*) = \mathrm{Ker}(\A)^{\perp} = \{u\in X'\simeq X:\int_{\Omega}uv = 0,\;\forall v\in \mathrm{Ker}(\A)\} $$ which is $\mathrm{span}\{1,x,x^2\}$.

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