Fina a singular value decomposition for $$A=\begin{bmatrix} -2 & 2 \\ -1 & 1 \\ 2 & -2\end{bmatrix}.$$ Find a (full svd)singular value decomposition for the matrix $A$.

My working thus far not sure how to determine u2 and u3 . A little confused on the whole process. ANy workings step by step that can assit would be appreciated. $$A^TA=\begin{pmatrix} 9 & -9 \\ -9 & 9 \end{pmatrix}$$
For the eigenvalue $\lambda=18$ I found a normalized eigenvector of $\frac{1}{\sqrt 2}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$. For $\lambda=0$ I found an eigenvector of $\frac{1}{\sqrt 2}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and so $$V=\frac{1}{\sqrt{2}}\begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}.$$ $\Sigma$ is the $3\times 2$ matrix whose diagonal is composed of the singular values $$\Sigma=\begin{pmatrix} \sqrt{18} & 0 \\ 0 & 0 \\ 0&0 \end{pmatrix}.$$ $$U1=\begin{pmatrix}\frac{2}{3} \frac{1}{3} \frac{-2}{3}\end{pmatrix}.$$

  • 1
    $\begingroup$ If your computations are correct, then any choice of $u_2$ and $u_3$ that makes $\{u_1, u_2, u_3\}$ into an orthonormal basis will suffice, since $u_2$ and $u_3$ disappear when you multiply by $\Sigma$. $\endgroup$ – angryavian Sep 7 '18 at 22:36
  • $\begingroup$ Yes your steps are correct .. bravo, what is your question now ? $\endgroup$ – Ahmad Bazzi Sep 7 '18 at 22:57

Your $A$ is rank $1$, as you can see by inspection. Thus the shape of $U,\Sigma,V^T$ in the reduced SVD are $3 \times 1,1 \times 1$ and $1 \times 2$ respectively. So you were actually done before computing the eigenvector with eigenvalue $0$.

If you want a full SVD, then you need shapes of $3 \times 3,3 \times 2$ and $2 \times 2$. To assemble this, you need to compute a unit eigenvector of $A$ with eigenvalue zero (which will be any unit vector orthogonal to $v_1$). This will be your $v_2$, which completes the set of right singular vectors. Then you need to get vectors $u_2,u_3$ that $\{ u_1,u_2,u_3 \}$ is an orthonormal system; this completes the set of left singular vectors.

  • $\begingroup$ Sorry Ian are you able to explain by workings still not getting it. $\endgroup$ – Liz Sep 8 '18 at 0:29
  • $\begingroup$ @Liz Do you want a full SVD or a reduced SVD? If you want a reduced SVD then you are already essentially done. $\Sigma=\begin{bmatrix} \sqrt{18} \end{bmatrix},V^T=\begin{bmatrix} -1/\sqrt{2} & 1/\sqrt{2} \end{bmatrix}$. To get $U$, multiply out $AV$ and normalize the columns (of which there is just one in this case). $\endgroup$ – Ian Sep 8 '18 at 2:12
  • $\begingroup$ @Liz But if you want a full SVD then you have some more calculations to do. $\endgroup$ – Ian Sep 8 '18 at 2:17

If $A$ is a matrix and $A \in \mathbb{R}^{ m \times n}$ then it has an SVD. i.e

$$A_{m \times n} = U_{m \times m} \Sigma_{m \times n} V_{n \times n}^{T} \tag{1}$$

now if the rank of $A$ is $r$ then the reduced or truncated SVD is given by an SVD like the following

$$ A_{m \times n} = U_{m \times r} \Sigma_{r \times r} V_{r \times n}^{T} \tag{2} $$

that means your reduced SVD is

$$ A_{m \times n} = \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{-2}{3} \end{bmatrix} \begin{bmatrix} \sqrt{18} \end{bmatrix} \begin{bmatrix} \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{bmatrix} \tag{3}$$

checking this in Python

import numpy as np

A1 = np.matrix([[-2, 2], [-1, 1], [2,-2]])

a1 = np.matrix([[2/3] , [1/3] ,[-2/3] ])
a2 = np.matrix([np.sqrt(18)])
a3 = np.matrix([-1/np.sqrt(2),1/np.sqrt(2)])

A2 = a1*a2*a3

error = np.linalg.norm(A1-A2)
Out[13]: 9.42055475210265e-16
  • $\begingroup$ Thanks RHowe. For u2 and u3 in part a) are you able to demonstrate how that comes about, $\endgroup$ – Liz Sep 8 '18 at 2:26
  • 1
    $\begingroup$ @Liz There is no $u_2,u_3$ in this answer; it's a reduced SVD of a rank 1 matrix, so only one of each type of singular vector is required $\endgroup$ – Ian Sep 8 '18 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.