Find SVD of a matrix

linear algebrasvd

Let A be a matrix, $A=
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 2 \\
\end{pmatrix}$ and it's SVD, $A=USV^t$.

Let $U^1$ be the first column of $U$ and $V^1$ the first column of $V$.

There are a bunch of options, but the correct answer is:

"The singular values of $A$ are $\sigma _1 = 2$ and $\sigma _2 =1$, $U^1=(0,1)$ and $V^1=(0,0,1)$"

And I think I get to the same matrix they're talking about, but in a different order?

  1. I found $A^tA$=$\begin{pmatrix}
    0 & 0 \\
    1 & 0 \\
    0 & 2 \\
    \end{pmatrix}\begin{pmatrix}
    0 & 1 & 0 \\
    0 & 0 & 2 \\
    \end{pmatrix}=\begin{pmatrix}
    0 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 4 \\
    \end{pmatrix}$
  2. I found the eigenvectors and eigenvalues of $A^tA$, $\lambda_1=0$, $\lambda_2=1$, $\lambda_3=4$. And their eigenvectors are $S_0=[(1,0,0)]$, $S_1=[(0,1,0])$, $S_4=[(0,0,1)]$.

I know I don't have to orthonormalize them, because they form the canonical basis.

So far, I have the singular values, which are $\sigma _1 = 1$ and $\sigma _2 =2$, $S=\begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
\end{pmatrix}$,$V=V^t=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}$

Now all I have to do is find $U$:

I find the columns for U:

$u_1=\frac{T(v_1)}{\sigma_1}=\frac{Av_1}{\sigma_1}=\frac{(1,0)}{1}=(1,0)$

$u_2=\frac{T(v_2)}{\sigma_2}=\frac{Av_2}{\sigma_2}=\frac{(0,2)}{2}=(0,1)$

Where $v_1$ and $v_2$ are the orthonormalized eigenvectors.

So $U$ must be $U=\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}$.

So I got that the singular values are $\sigma _1 = 1$ and $\sigma _2 =2$, $U^1=(1,0)$ and $V^1=(1,0,0)$ which are completly wrong by the correct answer I have.

What am I doing wrong?

Best Answer

I think you only used the wrong order. In the SVD one practically always arranges the singular values in descending order, so $\sigma_1 = 2$, $\sigma_2=1$. Therefore the diagonal matrix $S$ becomes $$ S = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} $$ and putting the corresponding vectors $v_1$, $v_2$, $v_3$ in the according order in a matrix gives $$ V = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}. $$ With this order, the vector $u_1$ becomes $(0,1)$.

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