$A=\begin{bmatrix} 1 & -3 & -3\\ 1 & 5 &1 \\ 1& 7 & 2 \end{bmatrix}$ , $\vec{b}=\begin{bmatrix} 5\\ -3\\ -5 \end{bmatrix}$
Part (c) of this question is what I am struggling with. Here is the rest of the question for context.
(a) Find an orthogonal basis for Col(A).
I have used the Gram-Schmidt process to create the set of vectors:
$\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ , $\begin{bmatrix} -6\\ 2\\ 4 \end{bmatrix}$ , $\begin{bmatrix} -18\\ 6\\ 12 \end{bmatrix}$.
(b) Find a basis for Nul(A)
I set up $A\vec{x}=0$, and got the following as the basis: $\begin{bmatrix} 3\\ 1\\ 0 \end{bmatrix}$, $\begin{bmatrix} 3\\ 0\\ 1 \end{bmatrix}$.
(c) Find the projection $\hat{b}$ of $\vec{b}$ onto Col(A).
Now, this is where I'm getting thrown off. Not because I do not understand how to do this calculation, but because I'm not sure what I am projecting onto. What I know is that Col(A) is the set of all linear combinations of the columns of $A$. So, at first glance I thought to project $\vec{b}$ onto the columns of $A$. However, part of me thinks I should project $\vec{b}$ onto the orthogonal basis for Col(A), because that was the first part of this question.
Am I overthinking what part (c) is asking?
Guide:
You have made a mistake earlier, if the basis in part $(a)$ has $3$ element and the basis in part $(b)$ has $2$ elements, this violates rank-nullity theorem which says that $$\operatorname{rank}(A)+\operatorname{nullity}(A) = 3$$
where $3$ is the number of columns.
Part $(a)$ should have $2$ elements in the basis and part $(b)$ should have $1$ element.
Orthogonal basis makes computation of projection easier. If the basis that you find in part $(a)$ is $\{ a_1, a_2\}$ To find projection $\vec{b}$ onto the the column space, just compute $\frac{\vec{b}^Ta_1}{\|a_1\|^2}(a_1) + \frac{\vec{b}^Ta_2}{\|a_2\|^2}(a_2)$
Alternatively, suppse the answer in part $(b)$ is $\{v\}$, then computing $\vec{b}-\frac{(\vec{b}^Tv)}{\|v\|^2}v$ works too.