Question

(1) A square matrix E є м,xn(R) is idempotent if E-E. It is symmetric if -t E. (a) Let V C Rn be a subspace of R, and consider the orthogonal projection projy R" Rn onto V. Show that the representing matrix E = projy18 of proj v relative to the standard basis of IRn is both idempotent and symmetric (b) Let E M/nxn(R) be a matrix that is both idempotent and symmetric. Show that there is a subspace V C Rn such that E-projylễ. [Hint: What does the hypothesis that E is symmetric say about the relation between ker E and im E?]

Please solve both parts of this question! I've stared at it for a long time without knowing how to approach it.

Answer 1

For further queries use comment section:

1.

Let  $\epsilon=\{e_1,e_2,\ldots ,e_n\}$ be  the standard basis vectors of.$\Bbb R^n$.

Let be generated by $\{e_1,e_2,\ldots ,e_m\}$ where .

Now $\text{proj}_V:\Bbb R^n\to \Bbb R^n$ is defined as

$\text{proj}_V(x)=\sum_{i=1}^m\dfrac{\langle x,e_i\rangle}{\langle e_i,e_i\rangle}e_i$

Hence

$\text{proj}_V(e_i)=e_i \forall 1\le i\le m\\ \& \\ \text{proj}_V(e_i)=e_i \forall m<i\le n\\$

Thus the matrix representing the above linear transform becomes

$E=\begin{bmatrix}\textbf{I}_{m\times m}&\textbf {0}\\\textbf {0}& \textbf {0}\end{bmatrix}$

Clearly $E=[\text{proj}_V]^\epsilon_\epsilon$ is a symmetric and idempotent.

Part-2:

Since $E^2=E$

so any vector $v\in \Bbb R^n$ can be uniquely represented as

v - Ev E ker E Ev ImageE

Notice that E is the identity operator on Image(E) since $E^2=E$

Let  

dim ImageE m

and let $\{Ev_i:1\le i\le m\}$ be a basis of .

ImageE

V, = span{Evi : 150 <m) S2

Now if we do a change of basis and since matrices are similar with respect to a change of basis we can find from $V^{'}$ with the required property.