SVD Singular Value Issue I am having an issue with a SVD problem and would like advice. The matrices given are
$$
  A= \left[ \begin{matrix}
    1 & 2\\
    3 & 6 \\
    \end{matrix}\right]\ \ \
  A^TA= \left[ \begin{matrix}
    10 & 20\\
    20 & 40 \\
    \end{matrix}\right]\ \ \
  AA^T= \left[ \begin{matrix}
    5 & 15\\
    15 & 45 \\
    \end{matrix}\right]
$$
I initially determined my $λ$ values to be $λ_1=50$ and $λ_2=0$ by computing the following:
$$
det⁡(A^TA-λI)=0\\
\begin{vmatrix}
    10-λ & 20\\
    20 & 40-λ \\
    \end{vmatrix}\\
(10-λ)(40-λ)-20^2=0\\
400-50λ+λ^2-400=0\\
λ^2-50λ=0\\
λ(λ-50)=0
$$
This in turn means that my singular values will be  $\sigma_1=\sqrt {50}$ and $\sigma_2=0$, and this is where I run into my issue. When I am trying to solve for the orthonormal set with $u_1$ and $u_2$, I was told to use the equation $u_n=\frac{Av_n}{\sigma_n}$. I cannot compute $u_2$ because that would make the denominator $0$, and the fraction would become undefined.
I discussed with my professor and they said I should try to calculate $u_2$ using the eigenvectors from $AA^T$, but $AA^T$ and $A^TA$ have the same eigenvalues. I thought maybe they meant to compute the $u$ for the $\sigma=\sqrt {50}$, but this is not the same answer as the one the book gives. The answers for this and Wolfram Alpha show that the SVD should be
$$
  \left[ \begin{matrix}
    1 & 2\\
    3 & 6 \\
    \end{matrix}\right]=
\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    1 & -3\\
    3 & 1 \\
    \end{matrix}\right]
  \left[ \begin{matrix}
    \sqrt{50} & 0\\
    0 & 0 \\
    \end{matrix}\right]
\frac{1}{\sqrt{5}}
  \left[ \begin{matrix}
    1 & -2\\
    2 & -1 \\
    \end{matrix}\right]
$$
I have gotten that $u_1=\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    1 \\
    3 \\
    \end{matrix}\right]$, but I do not know how to get $u_2=\frac{1}{\sqrt{10}}
  \left[ \begin{matrix}
    -3 \\
    1 \\
    \end{matrix}\right]$. Any help would be appreciated.
 A: Note that another way to write the singular value decomposition is to write it as the sum of rank one matrices:
$$A = U D V^T = \sum_{k = 1}^n \sigma_i \vec{u}_i \vec{v}_i^T$$
where $\vec{u}_i$ is the $i$th column of $U$ and $\vec{v}_i$ is the $i$th column of $V$. What does this tell us? It tells us that whenever $\sigma_i = 0$, the resulting left and right singular vectors don't really matter. So you have two choices:

*

*If $A$ is $m \times n$ (here $m = n = 2$) You can complete $\vec{u}_1$ to an orthonormal basis $\{\vec{u}_1, \vec{u}_2, \cdots, \vec{u}_m\}$ of $\mathbb{R}^m$ and complete $\vec{v}_1$ to an orthonormal basis $\{\vec{v}_1, \vec{v}_2, \cdots, \vec{v}_m\}$ of $\mathbb{R}^m$ of $\mathbb{R}^n$ (this can be done with Gram-Schmidt). As a result, $D$ will be an $m \times n$ diagonal matrix. That's what the solutions have done. Then we may write
$$A = \begin{bmatrix} \vec{u}_1 & \cdots & \vec{u}_m \end{bmatrix} \cdot \text{diag}_{m \times n}(\sigma_1, \sigma_2, \cdots, \sigma_{\min(m,n)}) \cdot \begin{bmatrix} \vec{v}_1 & \cdots & \vec{v}_n \end{bmatrix}^T$$


*Calculate the reduced singular value decomposition. This involves only worrying about the nonzero singular values, so we won't need to complete any set of singular vectors to a basis. Here $D$ will become an $r \times r$ diagonal matrix, where $r$ is the number of nonzero singular values ($r$ is also the rank of $A$). $U$ will be $n \times r$, and $V$ will be $m \times r$. In this case, we know $r = 1$ and you've calculated the corresponding left and right singular vectors of $\sigma_1 = \sqrt {50}$. So you can write
$$A = \frac{1}{\sqrt{10}}\begin{bmatrix} 1 \\ 3\end{bmatrix} \begin{bmatrix} \sqrt{50} \end{bmatrix} \left(\frac{1}{\sqrt 5} \begin{bmatrix} 1 & -2\end{bmatrix}\right)$$
Here $U$ and $V$ are partially orthogonal, in the sense that $U^T U = I_r = V^T V$ but $UU^T \neq I_n$ and $V V^T \neq I_m$.
A: Your professor was right.
Have you tried the definition of an eigenvector?
$$(AA^T)v=\lambda v$$
$$
\left[ \begin{matrix}
    5 & 15\\
    15 & 45 \\
\end{matrix}\right] 
\left[ \begin{matrix}
    v_1\\
    v_2\\
\end{matrix}\right] 
=\left[ \begin{matrix}
    0\\
    0\\
\end{matrix}\right] 
$$
$$
\begin{align}
& 5v_1+15v_2=0\\
& 15v_1+45v_2=3(5v_1+15v_2)=0
\end{align}
$$
So you have one degree of freedom to choose $v_2 = 1/\sqrt(10)$ and obtain $v_1 = -15/5 \, v_2 = -3/\sqrt(10)$.
