Consider the matrix:
$A= \begin{bmatrix} -1&1&0\\ -1&1&0\\ 1&1&0\\ 1&1&0 \end{bmatrix}$
Find the singular value decomposition (SVD). Then indicate the corresponding orthonormal basis for: $C(A)$,$N(A)$,$C(A^T)$, and $N(A^T).$
goal: $A = U \Sigma V^T$ U and V are orthogonal matrices. Their columns are orthonormal sets. $\Sigma$ is a diagonal matrix with non-negative entries.
Equations:
$(1) A^TA = V\Sigma^T\Sigma V^T \\(2) AV=U\Sigma$
Firstly, $A^TA = \begin{bmatrix} 4&0&0\\ 0&4&0\\ 0&0&0 \end{bmatrix}$
Compute eigenvalues: $\lambda_1 = 4 \lambda_2 = 4 \lambda_3 = 0$
The eigenvectors:
$v_1= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$,$v_2= \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$ $v_3 = \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$
Then, $V =\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}$
We get, $AV = \begin{bmatrix} -1&1&0\\ -1&1&0\\ 1&1&0\\ 1&1&0 \end{bmatrix}=U\Sigma $(This is $U\Sigma$ combined, to separate we need entries to have unit length).
$U=\begin{bmatrix} -\frac{1}{2}&\frac{1}{2}&0\\ -\frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}&0 \end{bmatrix}$ and $\Sigma$ is column space with square roots of the eigenvalues. $\Sigma =\begin{bmatrix} 2\\ 2\\ 0 \end{bmatrix}$.
Then here is my goal equation:$A = U \Sigma V^T$
$\Rightarrow A= \begin{bmatrix} -1&1&0\\ -1&1&0\\ 1&1&0\\ 1&1&0 \end{bmatrix} =\begin{bmatrix} -\frac{1}{2}&\frac{1}{2}&0\\ -\frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}&0 \end{bmatrix}\begin{bmatrix} 2\\ 2\\ 0 \end{bmatrix}\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}$
There must be an error. Not sure how to get the SVD.
Lastly,
$rref = \begin{bmatrix} 1&-1&0\\ 0&1&0\\ 0&0&0\\ 0&0&0 \end{bmatrix}$
The first column of U is spans the column space of A: $C(A) = \begin{bmatrix} -\frac{1}{2}\\ -\frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{bmatrix}$
The first column of V spans the row space of A: $C(A^T) =\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$
The second column of V spans the null space of A: $N(A) = \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$
The second and third columns of U span the left null space: $N(A^T)\{\begin{bmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{bmatrix},\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix} \} $