Activity 7.4.5

Suppose we have a singular value decomposition $A = U\Sigma V^T$ where $$U = \begin{bmatrix} \uvec_1 & \uvec_2 & \uvec_3 & \uvec_4 \end{bmatrix},\hspace{24pt} \Sigma = \begin{bmatrix} 18 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix},\hspace{24pt} V = \begin{bmatrix} \vvec_1 & \vvec_2 & \vvec_3 \end{bmatrix}\text{.}$$

1. What is the shape of $A$? What is $\rank(A)$?

2. Identify bases for $\col(A)$ and $\col(A^T)$.

3. Explain why $$U\Sigma = \begin{bmatrix} \uvec_1 & \uvec_2 \end{bmatrix} \begin{bmatrix} 18 & 0 & 0 \\ 0 & 4 & 0 \\ \end{bmatrix}\text{.}$$

4. Explain why $$\begin{bmatrix} 18 & 0 & 0 \\ 0 & 4 & 0 \\ \end{bmatrix}V^T = \begin{bmatrix} 18 & 0 \\ 0 & 4 \\ \end{bmatrix} \begin{bmatrix} \vvec_1 & \vvec_2 \end{bmatrix}^T\text{.}$$

5. If $A = U\Sigma V^T$, explain why $A=U_r\Sigma_rV_r^T$ where the columns of $U_r$ are an orthonormal basis for $\col(A)$, $\Sigma_r$ is a square, diagonal, invertible matrix, and the columns of $V_r$ form an orthonormal basis for $\col(A^T)$.