# Problem showing equality of original matrix to SVD

The SVD is defined as:

$$A=U\Sigma V^T$$

So, for each row of the matrix $a_i$, singular vector $u_i, v_i$ and singular value $\sigma_i$

$$a_i = \sigma_i u_i v_i^T$$

with
$$u_i = \frac{1}{\sigma_i}a_iv_i$$

so, my problem is the following....

$$a_i = \sigma_i\frac{1}{\sigma_i}a_i v_i v_i^T$$ It seems to me that: $$a_i \ne a_i v_i v_i^T$$

What's my mistake?

• If $u_i$ and $v_i$ are rows of the matrices $U$ and $V$, then $u_i v_i^T$ would be a $1\times n$ matrix multiplied by an $n\times 1$ matrix, resulting in a $1\times1$ matrix, not in an entire row. $\qquad$ – Michael Hardy Dec 23 '16 at 17:36

The mistake is in the first formula. You are saying that each row of $A$ depends only on one singular value, which is not true.
For instance let $$V=U=\frac1{\sqrt2}\begin{bmatrix}1&1\\1&-1\end{bmatrix},\ \ \ \Sigma=\begin{bmatrix}\sigma_1&0\\0&\sigma_2\end{bmatrix}.$$ Then $$A=\frac12\,\begin{bmatrix}\sigma_1+\sigma_2&\sigma_1-\sigma_2\\ \sigma_1-\sigma_2&\sigma_1+\sigma_2\end{bmatrix}.$$ You can check that your formula does not apply, as you could fix $\sigma_1$ and let $\sigma_2$ vary to make it fail.
$a_i = \sigma_i u_i v_i^T$ with $u_i = \dfrac{1}{\sigma_i}a_iv_i$