When I decompose a matrix $A$ using SVD method, how can I select $V$ in the following case?

Given $A=\begin{bmatrix}-1&2&2\end{bmatrix}$, decompose $A$ into $U \Sigma V^\text{T}$.

First, we know the size of $U$, $\Sigma$, and $V$ are $1\times1$, $1\times3$, and $3\times3$, respectively.

Second, we know that the columns of $U$ and $V$ are the eigenvectors of $AA^\text{T}$ and $A^\text{T}A$, respectively. Also, $UU^\text{T}=U^\text{T}U=I_{[1\times1]}$ and $VV^\text{T}=V^\text{T}V=I_{[3\times3]}$

Third, we know that the first $r$ diagonal components of $\Sigma$ is squared-root of the non-zero eigenvalues for both $A^\text{T}A$ and $AA^\text{T}$, and the rests are zero.

  1. We can caculate $$AA^\text{T} = \begin{bmatrix} 9 \end{bmatrix}\tag{1}\label{1}$$ and $$A^\text{T}A = \begin{bmatrix} 1&-2&-2\\ -2&4&4\\-2&4&4\end{bmatrix}.\tag{2}\label{2}$$

  2. Since the non-zero eigenvalues of both $AA^\text{T}$ and $A^\text{T}A$ is one and only one $9$. Hence, we have $$\Sigma = \begin{bmatrix} \sqrt9&0&0\end{bmatrix} = \begin{bmatrix} 3&0&0\end{bmatrix}.$$

  3. From \eqref{1}, we can have $$U = \begin{bmatrix} 1 \end{bmatrix}.$$

  4. From \eqref{2}, we can have one eigen vector corresponding to the eigenvalue $9$, which is $(-\frac13,~\frac23,~\frac23)$. However, I cannot determine other eigenvectors $(x,~y,~z)$ corresponding to the eigenvalue $0$, which should satisfy $x-2y-2z=0$ and $x^2+y^2+z^2=1$. Anyway, I can determine until $$V=\begin{bmatrix} -\frac13 & ? & ? \\ \frac23 & ? & ? \\ \frac23 & ? & ? \end{bmatrix}.$$

How can I determine $V$ in this case?

Actually, I can find $?$'s with an enough time, but I want to know easy way or formulaic way to find $?$'s.


The $9$-eigenvector is not correct. It should be $(-1,2,2)/3$ (or its negative).

For the other columns, note that $x-2y-2z=0$ determines a two-dimensional subspace spanned by $(2,1,0)$ and $(2,0,1)$. You can use Gram-Schmidt to find two orthonormal vectors for this subspace. I believe one such example is $(2,1,0)/\sqrt{5}$ and $(2, -4, 5)/5$.

  • $\begingroup$ Thank you. That is exactly my typo, sorry for perplex you. I correct from 1 to -1. Thank you for recalling the Gram-Schmidt method. Actually, I figured out the two orthonormal vectors using much time in the test. I should have known it can be done by Gram-Schmidt method. Thank you. $\endgroup$ – Danny_Kim Nov 21 '17 at 4:41

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