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.
Best Answer
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.