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Find the SVD form of the matrix $$A = \begin{pmatrix} 1 & 1 \\ 1& 1 \\ 1& 1 \end{pmatrix}$$

First, I calculated

$$A^TA = \begin{pmatrix} 3 & 3 \\ 3 & 3\end{pmatrix}$$

whose eigenvalues are $\lambda_1=6$, $\lambda_2=0$, so we get the singular values $\mu_1=\sqrt 6$ and $\mu_2=0$. Then the orthonormal eignvectors are: $u_1=1/\sqrt2 (1,1)$ And $u_2= 1/\sqrt2 (1,-1)$. So, I found the two matrix $\sum$ and $U^T$ where the SVD for A is of the form: $A=V \sum U^T$. I'm not sure how to find the orthogonal matrix V since I have a singular value which equals 0! Moreover i cannot multiply matrix because $\sum$ is not invertible. How can I finish it and get the SVD form for matrix $A$?

I am using information about SVD from the wikipedia page.

charmd
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Lam18373
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    You have a rank-1 matrix. Take a look at this. – Rodrigo de Azevedo Dec 03 '20 at 15:07
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    It's almost exactly the same thing. Only difference is that you have to find two vectors orthogonal to $(1,1,1)$ and to each other. This question can be solved without computing any eigenvalues. Give it another try. – Rodrigo de Azevedo Dec 03 '20 at 15:32
  • @Rodrigo de Azevedo I think i got it! So i wrote A as $(1,1,1)^T (1,1)= \sqrt(6) (1/√3,1/√3,1/√3)^T (1/√2,1/√2)$ that way I got $u_1, v_1$ and $\sigma_1$ then i can take any vectors $v_2$ orthonormal and orthogonal to $v_1$, $u_2,u_3$ orthonormal and orthogonal to $u_1$ .. there is a lot of choices in this method! Morover can i say that $\sigma_2=0$ since A of rank 1? – Lam18373 Dec 03 '20 at 18:48
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    Looks correct. Yes to all questions. Consider posting an answer to your own question. – Rodrigo de Azevedo Dec 03 '20 at 18:57
  • Okay, i'll try, do you know how to solve it by finding the vectors according to the SVD algorithm ? Which give us only specific solution? @Rodrigo de Azevedo – Lam18373 Dec 03 '20 at 19:00
  • On choices, take a look at this. – Rodrigo de Azevedo Dec 03 '20 at 19:27

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We can notice that we can write the matrix $A$ as follows:

$$A = \begin{bmatrix} 1 & 1 \\ 1 & 1\\ 1 & 1\end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix}^\top = \sqrt6\left(\frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right) \left( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right)^\top$$

So, we succeeded to write $A$ as $A=\sigma u_1 v_1$. In addition, $\sigma_2=0$ since $\mbox{rank}(A)=1$. Thus, we get

$$\Sigma= \begin{bmatrix} \sqrt{6} & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$$

Then in order to find the matrix $U$ and $V^T$ as in SVD A can be written as $A=U \Sigma V^T$. We can choose any vector $v_2 \in \Bbb R^2$ which is orthonormal and orthogonal to $v_1$ and choose vectors $u_2, u_3 \in \Bbb R^3$ which are also orthonormal and orthogonal to $u_1$.

Lam18373
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