Consider the matrix $A=\begin{bmatrix}1& \varepsilon\\ 0 & 0 \end{bmatrix}$ where $\varepsilon >0$.
Determine a singular value decomposition of $A$.
As far as I know the SVD can be calculated as follows:
$A=U\Sigma V^T$
in which $U$ are the eigenvectors of $AA^T$ and $V$ are the eigenvectors of $A^{T}A$.
The singular values are $0$ and $\sqrt{\varepsilon^2+1}\quad$
So $\Sigma=\begin{bmatrix}\sqrt{\varepsilon^2+1 }&0\\0& 0 \end{bmatrix}$
$AA^T=\begin{bmatrix}1+\varepsilon^2 & 0\\0&0\end{bmatrix}$ with eigenvalues $0$ and $1+\varepsilon^2$.
$A^TA=\begin{bmatrix}1&\varepsilon\\\varepsilon&\varepsilon^2 \end{bmatrix}$ with eigenvalues $0$ and $1+\varepsilon^2$.
In order to calculate the matix $U$, we take $(AA^T-\lambda I) \ \underline{u}_i = 0$.
For $\lambda = 1+\varepsilon^2$ this gives: $\begin{bmatrix}0&0\\0&-1-\varepsilon^2 \end{bmatrix} \underline{u}_1 = 0 \quad$ so $\underline{u}_1=\begin{bmatrix}1\\0 \end{bmatrix}$
For $\lambda=0$ this gives:
$\begin{bmatrix}1+\varepsilon^2&0\\0&0\end{bmatrix} \ \underline{u}_2=0 \quad$so $\underline{u}_2=\begin{bmatrix}0\\1 \end{bmatrix}$
So $U = \begin{bmatrix}1&0\\0&1\end{bmatrix}$
Next we calculate $V$: using $(A^TA-\lambda I)\underline{v}_i=0.$
For $\lambda=1+\varepsilon^2$ this gives: $\begin{bmatrix}-\varepsilon^2&\varepsilon\\\varepsilon&-1\end{bmatrix}\underline{v}_i = 0 \quad$ so $\underline{v}_1=\begin{bmatrix}\frac{1}{\varepsilon}\\1 \end{bmatrix}$
For $\lambda=0$ this gives:
$\begin{bmatrix}1&\varepsilon\\\varepsilon&\varepsilon^2\end{bmatrix} \ \underline{v}_2=0 \quad$so $\underline{v}_2=\begin{bmatrix}1\\ -\frac{1}{\varepsilon} \end{bmatrix}$
So $V=\begin{bmatrix}1&\frac{1}{\varepsilon}\\-\frac{1}{\varepsilon}&1 \end{bmatrix}$.
But if I now use $A=U\Sigma V^T$ I do not get the correct $A$.
I don't know what I'm doing wrong anymore.
Best Answer
$V$ contains two vectors the first is $v_1=[-\epsilon~~ 1]$ the second is $v_2=[1 ~~\epsilon]$