# SVD: How to find the columnvector of U corresponding to a singular value equal to zero

The question is if you have a situation where one of the singular values is equal to 0 in a singular value decomposition of a matrix, how to do you procede to find the column vector of U corresponding to this singular value?

Usually I found the columns of U with this relation, so let's say that we refer to $\vec{u_2}$ as the U column vector corresponding to singular value equal to 0:

$\vec{u_2} = (1/s_2)Av_2$

Now the as we can see the 0 is in an illegal place, since you can't divide by zero. But if you move it to the left side, I still can't quite see how to find the column of U corresponding to it. Any suggestions?

Here is an example of a matrix that provided this specific challenge

$A = \left(\begin{array}{rrr} -3 & 1 \\ 6 & -2 \\ 6 & -2 \end{array}\right).$

$\lambda{_1} = 90$,

$\lambda{_2} = 0$,

$s_1 = 3 \sqrt{10}$,

$s_2 = 0$,

How can one find $\vec{v_2}$ when $\lambda{_2} = 0$? In this case, you know that the eigenvectors of $A^{t} A$ are orthogonal, solve for $\lambda{_2}$ and find the eigenvector if it's orthogonal to the eigenvector corresponding to the first eigenvalue you have your solution.

So we end up with the following normalized column vectors for V, which we can see are orthogonal:

$\vec{v_1} = \left(\begin{array}{rrr} -3/\sqrt{10} \\ 1/\sqrt{10} \\ \end{array}\right).$

$\vec{v_2} = \left(\begin{array}{rrr} 1/\sqrt{10} \\ 3/\sqrt{10} \\ \end{array}\right).$

You can take any eigenvector for $v_2$. To find the columns of $U$ corresponding to $0$, you proceed as follows:
In this case, $AA^T=\begin{bmatrix} 10&-20&-20\\ -20&40&40\\ -20&40&40\\ \end{bmatrix}$. Row reducing, we get $\begin{bmatrix} 1&-2&-2\\ 0&0&0\\ 0&0&0\\ \end{bmatrix}$. So, the eigenvectors corresponding to $0$, say $\begin{bmatrix}x\\ y\\ z\end{bmatrix}$, satisfy the equation $x=2(y+z)$. So, the eigenvectors are given by $$\begin{bmatrix}x\\ y\\ z\end{bmatrix}= \begin{bmatrix}2(y+z)\\ y\\ z\end{bmatrix}=y\begin{bmatrix}2\\ 1\\ 0 \end{bmatrix}+ z\begin{bmatrix}2\\ 0\\ 1\end{bmatrix}$$ So, the eigenvectors corresponding to $0$ are $\begin{bmatrix}2\\ 1\\ 0 \end{bmatrix}$, $\begin{bmatrix}2\\ 0\\ 1\end{bmatrix}$. These are not orthogonal, so orthogonalise using Gram-Schmidt and then normalise. You will get the columns corresponding to $0$ for $U$.