We are given:
$$A = \begin{bmatrix}1 & 1\\1 & 0 \end{bmatrix}$$
We have:
$$W = A^T A = \begin{bmatrix}2 & 1\\1 & 1 \end{bmatrix}$$
The characteristic polynomial and eigenvalues of W are:
$$\lambda^2 +3 \lambda -1 = 0 \implies \lambda_1 = \frac{1}{2} \left(3-\sqrt{5}\right), ~ \lambda_2 = \frac{1}{2} \left(3+\sqrt{5}\right)$$
This gives us the singular values:
$$\sigma_1= \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)}, ~ \sigma_2= \sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)}$$
The eigenvectors of $W$ are:
$$v_1 = \begin{bmatrix} \frac{1}{2} \left(1-\sqrt{5}\right) \\ 1 \end{bmatrix}, ~ v_2 = \begin{bmatrix} \frac{1}{2} \left(1+\sqrt{5}\right) \\ 1 \end{bmatrix}$$
Normalizing these eigenvectors, we have:
$$v_1 = \begin{bmatrix} \frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \\ \frac{1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \end{bmatrix}, ~ v_2 = \begin{bmatrix} \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \\ \frac{1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \end{bmatrix}$$
Now, we can write $W = U \Sigma V^T$ as:
$$U = \left(
\begin{array}{cc}
\frac{\frac{1}{2} \left(1+\sqrt{5}\right)+1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1+\sqrt{5}\right)+1\right)^2}} & \frac{\frac{1}{2} \left(1-\sqrt{5}\right)+1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1-\sqrt{5}\right)+1\right)^2}} \\
\frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1+\sqrt{5}\right)+1\right)^2}} & \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1-\sqrt{5}\right)+1\right)^2}} \\
\end{array}
\right)$$
$$\Sigma = \left(
\begin{array}{cc}
\sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)} & 0 \\
0 & \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)} \\
\end{array}
\right)$$
$$V = \left(
\begin{array}{cc}
\frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} & \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \\
\frac{1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} & \frac{1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \\
\end{array}
\right)$$
Notes:
- Recall: The columns of V are called the right singular vectors. The columns of U are called the left singular vectors. To find $U$, we calculate the columns as:
$$u_1 = \dfrac{1}{\sigma_1} A v_1, ~ u_2 = \dfrac{1}{\sigma_2} A v_2$$
Care needs to be taken when one or both of the eigenvalues are zero!
This SVD can also be written as:
$$\sigma_1~u_1~v_1^T + \sigma_2~u_2~v_2^T$$
- Recall we need to form $V^T$ when writing this out.
Best Answer
In order to get the SVD of the matrix you need to get the eigenvalues of the covariance matrix
$$ A =\begin{bmatrix} 2& -6\\ 1 & -3 \\ 0 & 0 \end{bmatrix} \tag{1}$$
$$ A^{T}A = \begin{bmatrix} 5& -15\\ -15& 45 \end{bmatrix}\tag{2}$$
$$ \det(A^{T}A - \lambda I) = \begin{vmatrix} 5- \lambda& -15\\ -15& 45-\lambda \end{vmatrix} = (5-\lambda)(45-\lambda) -225 \tag{3}$$ $$ \det(A^{T}A - \lambda I) = \lambda(\lambda-50) \implies \lambda_{1} = 50 \lambda_{2} = 0 \tag{4}$$
now $A^{T}A = V \Lambda V^{T} $
we find the right singular vectors
$$ A^{T}A -50I = \begin{bmatrix} 5-50& -15\\ -15& 45 -50 \end{bmatrix} \begin{bmatrix}x_{1} \\ x_{2} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} \tag{5}$$
$$ A^{T}A-50I = \begin{bmatrix} -45 &-15\\ -15& -5 \end{bmatrix} \begin{bmatrix}x_{1} \\ x_{2} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} \tag{6}$$
$$ -45x_{1} -15x_{2} = 0 \\ -15x_{1} - 5x_{2} = 0 $$ then $x_{2} = \frac{x_{1}}{3}$
$$ x = \begin{bmatrix} 1 \\ 3\end{bmatrix} \tag{7} $$
now normalize
$$ v_{1} = \begin{bmatrix} \frac{1}{\sqrt{3^{2}+1}}\\ \frac{3}{\sqrt{3^{2}+1}} \end{bmatrix} \tag{8} $$ $$ v_{1} = \begin{bmatrix} \frac{1}{\sqrt{10}}\\ \frac{3}{\sqrt{10}} \end{bmatrix} \tag{9} $$
for the next
$$ A^{T}A = \begin{bmatrix} 5& -15\\ -15& 45 \end{bmatrix} \begin{bmatrix}x_{1} \\ x_{2} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} \tag{10}$$
you get the opposite
$$ v_{2} = \begin{bmatrix} \frac{-3}{\sqrt{10}}\\ \frac{1}{\sqrt{10}} \end{bmatrix} \tag{11} $$
the singular values are the square roots of the eigenvalues so..you have $\sigma_{1} = \sqrt{50} , \sigma_{2} =0$
if you do it by python ..
import numpy as np
these aren't unique.. you should note..
if you do the $AA^{T}$
$$AA^{T} = \begin{bmatrix} 40& 20 & 0 \\ 20 & 10 & 0 \\ 0 & 0 & 0 \end{bmatrix} \tag{12}$$
then take
$$ \det(AA^{T} - \lambda I) = \begin{vmatrix} 40-\lambda& 20 & 0 \\ 20 & 10-\lambda & 0 \\ 0 & 0 & 0 \end{vmatrix} \tag{13}$$
in the end if $U \Sigma V^{T} = A $ you'll be fine..