# Eigenvalues of Orthogonal Projection, using representative matrix

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 $$$$\left[Proj_{\;U} \right]_{B_{v}} = \begin{bmatrix}\dfrac{\langle v_1,u_1 \rangle}{||u_1||^2} &\ldots &\dfrac{\langle v_n,u_1\rangle}{||u_1||^2}\\ \vdots&\ddots&\vdots\\ \dfrac{\langle v_1,u_m \rangle}{||u_m||^2}&\ldots& {\dfrac{\langle{v_n,u_m}\rangle}{||u_m||^2}} \end{bmatrix}$$$$ This doesn't convince me.
is this wrong? How can i find the eigenvalues of this transformation? Thanks in advance-

• Hint: Find a basis of $V$ in which the matrix is diagonal. – amd Dec 30 '18 at 21:39

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$$\begin{bmatrix}\operatorname{Id}_m&0\\0&0_l\end{bmatrix}.$$