# Determine orthogonal projection on column space of given matrix

I have been given this problem to solve for my next lecture. I tried to solve it and this is what i came up with. I need someone to check this, is it correct answer.

Problem: Find linear operator of orthogonal projection $$P:R^{3}->R^{3}$$ which projects onto column space of matrix B. $$B = \bigl(\begin{smallmatrix} 1 &1 \\ 0&1 \\ 2&0 \end{smallmatrix}\bigr)$$

a)Find matrix of linear operator P, in the base of space R3. Choose basis vectors as you want.
b) Find image of vector b=[-4 3 4]^t with P
c) Without doing any calculation find eiganvalues and eigenvectors for P

Here is my atempt to solve this problem.

a) We know that linear operator P projects onto column space of B, so $$Im(P)=Lin{ { \begin{bmatrix} 1\\ 0\\ 2 \end{bmatrix}, \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} } }$$ For the kernel of P. I think that every other vector should be projected to zero. So i found a orthogonal subspace spanned by vector $$\begin{bmatrix} -2\\ 2\\ 1 \end{bmatrix}$$ So $$KerP= Lin \begin{bmatrix} -2\\ 2\\ 1 \end{bmatrix}$$

If i choose these vectors to be basis for R3 and represent the P by this base it's matrix will be:

$$P= \begin{matrix} 1 &0 &0 \\ 0&1 & 0\\ 0 & 0& 0 \end{matrix}$$

b) P*b c) Obviously the eigenvalues are 1,1 and 0. And the eigenvectors are easly calculated without touching the pen it's the standard (canonical) base for R3.

I don't think the point of the problem is that you can claim part b is stated relative to your chosen basis, nor part c. To get part b you would have to express vector $$\vec b$$, assumed relative to the standard basis for $$\mathbb{R}^3$$ in terms of your basis, project with your $$P$$, then express the result in terms of the original basis. Thus if $$V=\begin{bmatrix}1&1&-2\\0&1&2\\2&0&1\end{bmatrix}$$ Is the matrix whose columns are your basis vectors, then you need to compute \begin{align}VPV^{-1}\vec b&=\begin{bmatrix}1&1&-2\\0&1&2\\2&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}\begin{bmatrix}\frac19&-\frac19&\frac49\\\frac49&\frac59&-\frac29\\-\frac29&\frac29&\frac19\end{bmatrix}\vec b\\ &=\begin{bmatrix}\frac59&\frac49&\frac29\\\frac49&\frac59&-\frac29\\\frac29&-\frac29&\frac89\end{bmatrix}\begin{bmatrix}-4\\3\\4\end{bmatrix}=\begin{bmatrix}0\\-1\\2\end{bmatrix}\end{align} to get the projection in part b. Note that as an intermediate result in our calculation we found $$P$$ relative to the standard basis for $$\mathbb{R}^3$$.
The columns of $$V$$ above are eigenvectors of $$P$$ with eigenvalues $$1$$, $$1$$, and $$0$$ relative to the standard basis for $$\mathbb{R}^3$$ by construction.
• What is your [numeric] answer to part b? If you said $$\begin{bmatrix}-4\\3\\0\end{bmatrix}$$ the answer clearly wouldn't be right in either basis. – user5713492 Jun 12 at 19:00
• I suppose the solution in your cross comment would be OK, with an explanation that the answer is relative to your new basis. That is pretty much what my arithmetic did except that it then transformed back to the standard basis for $\mathbb{R}^3$. – user5713492 Jun 12 at 19:03