I've encountered the following question:
Assume that we have \begin{equation*} A = \begin{pmatrix} 1 & -1 & 1\\ -1 & 1 & 1 \\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0\\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 & 1\\ 1 & 2 & 0 \\ -2 & 1 & 5 \end{pmatrix} \end{equation*}
Now, label the columns of leftmost matrix as $a_{1}, a_{2}$ and $a_{3}$, the diagonal entries of the middle matrix by $t_{1}, t_{2}$ and $t_{3}$, and finally the rows of the rightmost matrix by $b_{1}^{T}, b_{2}^{T}$ and $b_{3}^{T}$. The question is:
Question: Show how to modify the $a_{i}, b_{i}$ and $t_{i}$'s to yield a singular value decomposition for $A$.
Here, I can write $A$ as sum of rank-1 matrices using $a_{i}, b_{i}$ and $t_{i}$'s (which was asked in the latter part of this question). But I don't know what the question mean by modifying $a_{i}, b_{i}$ and $t_{i}$'s? Can anyone help?
Thanks in advance.
Let us take the following notations:
$$M= \underbrace{\begin{pmatrix} 1 & -1 & 1\\ -1 & 1 & 1 \\ 2 & 1 & 0 \end{pmatrix}}_A \underbrace{\begin{pmatrix} 1 & 0 & 0\\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{pmatrix}}_T \underbrace{\begin{pmatrix} 2 & -1 & 1\\ 1 & 2 & 0 \\ -2 & 1 & 5 \end{pmatrix}}_B$$
$A$ has orthogonal (but not normalized) columns.
$B$ has orthogonal (but not normalized) lines.
$T$ is diagonal but one its entries is negative (instead of all positive).
These deficiencies can be compensated by inserting diagonal matrices in the following way:
$$M=A\underbrace{\begin{pmatrix} 1/\sqrt{6} & 0 & 0\\ 0 & 1/\sqrt{3} & 0 \\ 0 & 0 & 1/\sqrt{2} \end{pmatrix}}_{D_1} D_2D_2 T \underbrace{\begin{pmatrix} 1/\sqrt{6} & 0 & 0\\ 0 & 1/\sqrt{3} & 0 \\ 0 & 0 & 1/\sqrt{30} \end{pmatrix}}_{D_3}B$$
with $$D_2:=\begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
Setting $$U=AD_1D_2, \ \ \ \Sigma=D_2T, \ \ \ V^T=D_3B,$$ we recover the SVD of $M$:
$$M=U\Sigma V^T$$