Question

Suppose that {ūj, ..., ūk} is an orthonormal basis for a subspace W of R" and we form the matrix U = (ū; ū2 ... ük) Then the matrix P= UUT has the property that p2 = P . This follows for the following reason(s). A. We know that P= I and so P2 = 1? = I = P

Answer 1

The basis
{u1, ..., ик
of is orthonormal, i.e., $u_i^Tu_j=\delta_{ij}$ where $\delta_{ij}=1$ if and only if
2=
, otherwise $\delta_{ij}=0$ .

It is not necessary that is the identity matrix. For example, let be the subspace spanned by the line $y=x$ in $\mathbb{R}^2$ . Then

$u_1=\begin{bmatrix} \tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} \end{bmatrix}$

is a basis of and hence

$P=UU^T=\begin{bmatrix} \tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} \end{bmatrix}\begin{bmatrix} \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \end{bmatrix}=\begin{bmatrix} \tfrac{1}{2} & \tfrac{1}{2}\\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}$

is not identity. Thus option A. is incorrect.

Option B. is correct. Since we $u_i^Tu_j=\delta_{ij}$ , implies $U^TU=I$ and thus $P^2=P$

Option C. is also correct.

The projection matrix onto a subspace is given by

$U(U^TU)^{-1}U^T$

where $U=\begin{bmatrix} u_1& \ldots& u_k \end{bmatrix}$ and
{u1, ..., ик
is a basis of .

Since
{u1, ..., ик
is an orthonormal basis, we have $U^TU=I$ and hence the projection matrix is given by $U(U^TU)^{-1}U^T=UU^T$ .

Thus $P=UU^T$ is a projection matrix and hence $P^2=P$

Thus the correct option is option E, i.e., both B and C are correct.