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I am learning about SVD and some basic calculations regarding that topic. Here is a small exercise that I am trying to do, that is, to find the SVD of

$$A=\begin{bmatrix} -1 & 1\\1 & 1\\1 & 2\end{bmatrix}$$

Here is a summary of my steps and findings:

(1.a) Find $A^\top A$.

(1.b) Find its eigenvalues. Found to be $7, 2$

(1.c) Find corresponding normalized eigenvectors and construct matrix $V$ with above eigenvectors as its columns. That is,

$$ V = \begin{bmatrix}\frac{1}{\sqrt{5}}& -\frac{2}{\sqrt{5}}\\\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{bmatrix}$$

(2) Construct $S$ same order as of $A$ with square root of eigenvalues along its diagonal. That is,

$$ S = \begin{bmatrix}\sqrt{7} & 0\\0 &\sqrt{2}\\0 & 0\end{bmatrix} $$

(3.a) Find $A A^\top$

(3.b) Find its eigenvalues. They would be $7, 2$ [directly from (1.b)] and $0$ [Number of rows - number of columns = 1]

(3.c) Find corresponding normalized eigenvectors. Construct matrix $U$ with above eigenvectors as its columns. That is, $$ U = \begin{bmatrix} \frac{1}{\sqrt{35}} & -\frac{3}{\sqrt{10}} & -\frac{1}{\sqrt{14}} \\ \frac{3}{\sqrt{35}} & \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{14}} \\ \frac{5}{\sqrt{35}} & 0 & \frac{2}{\sqrt{14}} \end{bmatrix}$$

Then $A = U S V^\top$ but after checking I found that $A \ne U S V^\top$. Though if we interchange the signs of the second column of $V$, that is, just taking the negative of the eigenvector (which is also an eigenvector of $A^\top A$ corresponding to eigenvalue $2$), then we indeed get $A=USV^t$

Question: what went wrong in my steps?

Rodrigo de Azevedo
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Savitr
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  • See my post here. Does that clear things up? – Ben Grossmann Mar 21 '25 at 15:39
  • @BenGrossmann Thank you. I read a little more of the proof in the text about existence of SVD and it used that the some of the column vectors of U are related to the column vectors of V. It was confusing since in the example they did not use that and only calculated U And V from eigenvectors of AA* and AA. So my goal should be find normalized eigenvectors {v_1, v_2, ..., v_n} of AA corresponding to non-decreasing eigenvalues, that will give V. S is obtained from taking positive square roots of eigenvalues and arranging them diagonally and filling other elements by 0. – Savitr Mar 21 '25 at 16:31
  • If only first $r$ eigenvalues are non-zero, then we normalize ${Av_1, \cdots , Av_r}$, and extend it to a orthonormal basis for $\mathbb{C}^m$ to get $U$ – Savitr Mar 21 '25 at 16:35
  • That's exactly right – Ben Grossmann Mar 21 '25 at 16:36
  • And in case $m<n$, it may be easier to calculate eigenvalues from $AA^\ast$ and get corresponding eigenvectors to find $U$. And then we can get $v_i=\frac{1}{||A^\ast u_i||}(A^\ast u_i)$ and extend it. – Savitr Mar 21 '25 at 16:44
  • Right again. Equivalently, you can find the SVD of $A^$, and then use the fact that $$ (U \Sigma V^)^* = V\Sigma^U^ $$ – Ben Grossmann Mar 21 '25 at 16:45
  • Related – Rodrigo de Azevedo Mar 22 '25 at 13:05

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