Let $V$ an inner product vector space and $U$ a vector subspace of $V$. Consider the linear operator $Proj_{\; U }:V\rightarrow V$ such that $\forall v\in V \; : \; Proj_{\; U}(v) = Proj_{\; U}V$, where $Proj_{\; U}V$ is the orthogonal projection of the vector $v$ onto the subspace $U$. Find the representative matrix of $Proj_{\; U}V$ and show that it's eigenvalues are $\lambda_1= 0$ and $\lambda_2 = 1$.
I've tried to define $B_v =\left\{v_1,\ldots,v_n \right\} $ as a basis of $V$ and $B_u =\left\{u_1,\ldots,u_m \right\} $ as an orthonormal basis of $U$. Then i got the matrix \begin{equation}\left[Proj_{\;U} \right]_{B_{v}} = \begin{bmatrix}\dfrac{\langle v_1,u_1 \rangle}{||u_1||^2} &\ldots &\dfrac{\langle v_n,u_1\rangle}{||u_1||^2}\\ \vdots&\ddots&\vdots\\ \dfrac{\langle v_1,u_m \rangle}{||u_m||^2}&\ldots& {\dfrac{\langle{v_n,u_m}\rangle}{||u_m||^2}} \end{bmatrix} \end{equation}
This doesn't convince me.
is this wrong? How can i find the eigenvalues of this transformation?
Thanks in advance-
As you wrote, let $\{u_1,\ldots,u_m\}$ be an orthonormal basis if $U$. Add vectors $v_1,\ldots,v_l$ to it so that $B=\{u_1,\ldots,u_m,v_1,\ldots,v_l\}$ is an orthonormal basis of $V$. Then the matrix of $\operatorname{Proj}_U$ with respect to this basis is$$\begin{bmatrix}\operatorname{Id}_m&0\\0&0_l\end{bmatrix}.$$