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\begin{align} A & =\begin{bmatrix}2&3\\6&9\end{bmatrix} \end{align} I found four fundamental subspaces of A which are

\begin{align} C(A) & =\begin{bmatrix}2\\6\end{bmatrix} , dimension = 1 \end{align}

\begin{align} C(A^T) & =\begin{bmatrix}2\\3\end{bmatrix} , dimension = 1 \end{align}

\begin{align} N(A) & =\begin{bmatrix}-3\\2\end{bmatrix} , dimension = 1 \end{align}

\begin{align} N(A^T) & =\begin{bmatrix}3\\-1\end{bmatrix} , dimension = 1 \end{align}

The first question is that "Find the projection matrices onto each of the fundamental subspaces" and I used the formula that the projection matrix on to C(A) is
\begin{align} P & = A(A^TA)^{-1}A^T \end{align}

However, how to find the projection matrix if

\begin{align} det(A^TA) = 0 ? \end{align} I thought I couldn't find the projection matrix onto C(A) and even other subspaces because A^TA is singular matrix.

So, please let me know what is projection matrix onto each of the fundamental subsapces and how to find them.

Next question is that "Write the vector b = (2,2) as a linear combination of a vector in C(A) and N(A^T)

I thought this question was asking to solve

\begin{align} a & \begin{bmatrix}2\\6\end{bmatrix} + b\begin{bmatrix}3\\-1\end{bmatrix} = \begin{bmatrix}2\\2\end{bmatrix} \end{align}

using vectors from C(A) and N(A^T) we found in previous problem. so, I got a = 2/5 and b = 2/5 from the above equation I made.

Please let me know how to write vector b as a linear combination of a vector in column space and left null space.

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  • $\begingroup$ That formula only applies when $A$ has full rank. Otherwise, as you’ve discovered, $A^TA$ is not invertible. $\endgroup$ Commented Dec 8, 2018 at 7:05

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Projection of a vector $\bf{u}$ along the vector $\bf{v}$ is given by $\textbf{proj}_{\bf{v}}\bf{u}=\left(\frac{\bf{u} \cdot \bf{v}}{\bf{v} \cdot \bf{v}}\right)\bf{v}$. So to get the projection matrix we can see what happens to the standard basis vectors $\bf{e}_1$ and $\bf{e}_2$, i.e. $$\textbf{proj}_{\bf{v}}\bf{e_1}=\left(\frac{\bf{e}_1 \cdot \bf{v}}{\bf{v} \cdot \bf{v}}\right)\bf{v} \qquad \textbf{proj}_{\bf{v}}\bf{e_2}=\left(\frac{\bf{e}_2 \cdot \bf{v}}{\bf{v} \cdot \bf{v}}\right)\bf{v}$$ Suppose we want the projection matrix for the fundamental space $C(A^T)$ so $\bf{v}=\begin{bmatrix}2\\3\end{bmatrix}$. Then, $$\textbf{proj}_{\bf{v}}\bf{e_1}=\left(\mathrm{\frac{2}{13}}\right)\bf{v} \qquad \textbf{proj}_{\bf{v}}\bf{e_2}=\left(\mathrm{\frac{3}{13}}\right)\bf{v}.$$ The projection matrix is given by $$P=\begin{bmatrix}\uparrow & \uparrow\\ \textbf{proj}_{\bf{v}}\bf{e_1} & \textbf{proj}_{\bf{v}}\bf{e_2}\\ \downarrow & \downarrow \end{bmatrix}=\begin{bmatrix}\frac{4}{13} & \frac{6}{13}\\\frac{6}{13} & \frac{9}{13}\end{bmatrix}$$ Now you can compute other projection matrices as well.

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