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.
Best Answer
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 $u_1$.