# uniqueness of svd decomposition and its role in statistical analysis

let us consider following model

http://www4.ncsu.edu/~ipsen/REU09/chapter4.pdf

it says that :

The singular values are unique, but the singular vector matrices are not

it was shown also in this simulink simulation,in matlab command place i have got following result

A=[2 1 3;1 4 5]

A =

2     1     3
1     4     5

>> [U E V]=svd(A)

U =

-0.4719   -0.8817
-0.8817    0.4719

E =

7.2965         0         0
0    1.6617         0

V =

-0.2502   -0.7772   -0.5774
-0.5480    0.6053   -0.5774
-0.7982   -0.1720    0.5774


simulink shows me same singular values,but singular vectors are different in sign compare to calculation by hand.now if i will use result given by simulink in some statistical analysis,does it give me different result then usage of result given by hand calculation? thanks in advance

The singular vectors may differ in sign, as the algorithm used to compute them is iterative and there is no guarantee that you will get the singular values in the same order, or if their sign is different. Note that the decomposition $A = U\Sigma V^*$ will be the same if we multiply a factor of $-1$ to $U$ and $V$:

$$A = U\Sigma V^* = (-1)^2U \Sigma V^* = (-U)\Sigma (-V)^*.$$

Nevertheless, this should not affect your use of these singular vectors, as they still span the same vector space regardless of the parity.

The freedom of signs (or complex phase) is detailed in Calculating SVD by hand.

The crux of the problem is summarized in this example where $$\mathbf{A} = \left[ \begin{array}{cc} u_{1} & u_{2} \end{array} \right] % \Sigma \, % \left[ \begin{array}{ccc} v_{1} & v_{2} & v_{3} \end{array} \right]^{*}$$ If we select the sign convention $\color{blue}{\pm}$ on the $k-$th column vector in the range space, we induce the sign convention $\color{red}{\pm}$ on the $k-$th column vector in the complimentary range space. The $\color{blue}{choice}$ induces the $\color{red}{consequence}$.

For example, find the SVD by resolving the column space to produce the vectors $u$. The vectors $v$ will be constructed using the second equality in $(1)$. Flipping the global sign on $u_{1}$ flips the global sign on $v_{1}$. $$\mathbf{A} = \left[ \begin{array}{cc} \color{blue}{\pm}u_{1} & u_{2} \end{array} \right] % \Sigma \, % \left[ \begin{array}{ccc} \color{red}{\pm}v_{1} & v_{2} & v_{3} \end{array} \right]^{*}$$

The number of unique sign choices is $2^\rho$, two choices for each range space vector.

Here the matrix rank is $\rho = 2$ and there are $2^{2} = 4$ unique vector sets. If $\mathbf{A}\mathbf{A}^{*}$ is resolved, then: $$\begin{array}{cc|cc} u_{1} & u_{2} & v_{1} & v_{2} \\\hline \color{blue}{+} & \color{blue}{+} & \color{red}{+} & \color{red}{+} \\ \color{blue}{+} & \color{blue}{-} & \color{red}{+} & \color{red}{-} \\ \color{blue}{-} & \color{blue}{+} & \color{red}{+} & \color{red}{+} \\ \color{blue}{-} & \color{blue}{-} & \color{red}{-} & \color{red}{-} \\ \end{array}$$