Finding a subspace of orthogonal projection

Question

I would like some help solving this question

P is an orthogonal projection that is defined by:

$$P:\Bbb R^3\to \Bbb R^3$$ $$P(v)=1/180\begin{pmatrix}144&72&0\\72&36&0\\ 0&0&180\end{pmatrix}·v$$ How do I find subspace $W$ so that $$P=P_w, \quad ?$$ I've searched everywhere online but i can't seem to find a solution to it.

Answer 1

Let $A$ the associated matrix.

We verify that $A^2=A$.

The null space of A is $\text{span}\left\{\begin{pmatrix}-10\\20\\0\end{pmatrix}\right\}$

The column space of A is $\text{span}\left\{\begin{pmatrix}0\\0\\1\end{pmatrix}, \begin{pmatrix}4/5\\2/5\\0\end{pmatrix}\right\}$

They are orthogonal.

P is the orthogonal projection on $W=\text{span}\left\{(0,0,1),(4/5,2/5,0)\right\}$

Answer 2

First, I checked and, yes, the determinant of that matrix is 0 so it does map R^3 onto a proper subspace of dimension 2, or 1. Let $v= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$ be a vector in $R^3$.

Then $Pv= \frac{1}{180}\begin{bmatrix}144 & 72 & 0 \\ 72 & 36 & 0 \\ 0 & 0 180 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \frac{1}{180}\begin{bmatrix}144x+ 72y \\ 72x+ 36y \\ 180z\end{bmatrix}$

If x= y= 0 and z= 1, that is $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ so that is one basis vector.

If x= z= 0 and y= 1, that is $\begin{bmatrix}\frac{72}{180} \\ \frac{36}{180} \\ 0 \end{bmatrix}= \begin{bmatrix}\frac{2}{5} \\ \frac{1}{5} \\ 0 \end{bmatrix}$. That is not a multiple of $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$ so it is another basis vector.

We can now say that, since the maximum dimension of the subspace was 2, these two vectors form a basis.

If you were wondering, yes, we could also take y= z= 0 and x= 1 to get $\begin{bmatrix}\frac{144}{180} \\ \frac{72}{180} \\ 0 \end{bmatrix}= \begin{bmatrix}\frac{4}{5} \\ \frac{2}{5} \\ 0 \end{bmatrix}$. But $\begin{bmatrix}\frac{4}{5} \\ \frac{2}{5} \\ 0 \end{bmatrix}= 2 \begin{bmatrix}\frac{2}{5} \\ \frac{1}{5} \\ 0 \end{bmatrix}$ so that is not an independent vector.