# Find the singular value decomposition.

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: $N(A^T)\{\begin{bmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{bmatrix},\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix} \}$

You have written $\Sigma$ as a vector. It is a diagonal matrix. The rank of $A$ is two (why?). Thus, column space and row-space should have two vectors in their basis.