# Argue that the transformation is diagonalizable.

Let V be an (n−1)-dimensional subspace of $\mathbb R^n$. Argue that the transformation f : $\mathbb R^n → \mathbb R^n$ which orthogonally projects a vector onto V is diagonalizable. Moreover, compute the eigenvalues of f, the trace of f and the determinant of f.

For the first part, I know that if V is diagonalizable then there exists a D where the diagonal are the eigenvalues. I also know that in order for it to be diagonalizable it would have to have the same number of eigenvectors as the dimension. The problem is that all I'm given is the fact that it orthogonally projects a vector onto V. So given that information, how would I be able to argue that it's diagonalizable?

For second part, I know that the trace is the sum of its eigenvalues and the determinant is it's product, but given that f orthogonally projects a vector onto V, would that change it?

By the definition of the orthogonal projection, if $v \in V$ then $f(v) = v$ and if $v \in V^{\perp}$ then $f(v) = 0$. Try to justify why you can find a basis $(v_1,\dots,v_n)$ of $\mathbb{R}^n$ such that $v_1,\dots,v_{n-1} \in V$ and $v_n \in V^{\perp}$. Then, we will have $f(v_i) = v_i$ for $1 \leq i \leq n - 1$ and $f(v_n) = 0$ which shows that each $v_i$ is an eigenvector of $f$ and so $f$ is diagonalizable. The corresponding eigenvalues are $1$ (with multiplicity $n - 1$) and $0$ (with multiplicity $1$) and so $f$ is diagonalizable and
$$\det(f) = 1 \cdot (\ldots) \cdot 1 \cdot 0 = 0, \\ \operatorname{tr}(f) = 1 + \dots + 1 + 0 = n - 1.$$