I have a matrix like $\mathbf{L} = \begin{bmatrix}2&0&4\\0.5&1&0\end{bmatrix}$ And I want to find out the image space and null space of it.
What I did is like , I reduced the rows to \begin{bmatrix}1&0&2\\0&1&-1\end{bmatrix}. From this I got to know that the null space can be seen in the free variable column. Thus null space is $$c\cdot\begin{bmatrix}-2\\1\\1\end{bmatrix}$$.
- What can I say about the column space of the matrix.? I know that the column space is the linear combination of the columns of L
- When I find the
[U,S,V] = svd(L)using Matlab. The matrices I am getting is, $$U=\begin{bmatrix}-0.9986&0.0531 \\-0.0531&-0.9986\end{bmatrix}$$ $$S=\begin{bmatrix}4.4781&0&0 \\0&1.0940&0\end{bmatrix}$$ $$V=\begin{bmatrix} -0.4519&-0.3593 &-0.8165\\ -0.0119 & -0.9128 & 0.4082\\ -0.8920 & 0.1942 & 0.4082 \end{bmatrix}$$
I know that the last column of th V matrix corresponds to the null space of the matrix. $$x_n = \begin{bmatrix} -0.8165\\ 0.4082\\ 0.4082 \end{bmatrix} \Longleftrightarrow \begin{bmatrix} \frac{-2}{\sqrt{6}}\\ \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{6}} \end{bmatrix}$$
How can I get to know what are the image space of the matrix , and what to be done to get the image space came out of svd, from the normal image space means the linear combination of columns of $L$