# Image and Kernel of a matrix using SVD

In this post kernel and image have been discussed. The theory is not really tackled, but it is a good example which fits the theory which one can for example read in wikipedia.

However, I'd like to know how this can be achieved easier if the SVD is given. I have given a concrete exercise:

The controllability matrix $$P=\left[\begin{array}{lll}{A^{2} B} & {A B} & {B}\end{array}\right]$$ of a state-space model $$(A,B,C,D)$$ has the following SVD. Determine the kernel and image space of $$P$$. $$P=\left[\begin{array}{lll}{u_{1}} & {u_{2}} & {u_{3}}\end{array}\right]\left[\begin{array}{ccc}{\sigma_{1}} & {0} & {0} \\ {0} & {\sigma_{2}} & {0} \\ {0} & {0} & {\sigma_{3}}\end{array}\right]\left[\begin{array}{c}{v_{1}^{T}} \\ {v_{2}^{T}} \\ {v_{3}^{T}}\end{array}\right]$$

For the kernel I have to find something that maps A to $$0$$ --> $$Ax=0$$. U and V are used for scaling and rotation so I assume it has something to do with $$\Sigma$$.

For the image I have trouble understanding the definition. Neither in my old scripts nor in Wikipedia I have found any examples. Only definitions of what it is supposed to be. I failed to connect the example from the added post to the theory. Therefore I also do not see how SVD helps me there.

Does someone see how the given SVD can be beneficial? It has to be quite simple given the amount of points which were given for this exercise.

EDIT: Just found this:

The SVD will give us the rank of matrix A by simply getting the number of nonzero singular values of A or the nonzero diagonal elements of Σ.

I assume this is not the final solution but the rank is quite often mentioned in this whole kernel/image theory. The quote makes sense. I see why this is the rank. Still don't see how that leads to kern() and Im().

EDIT2: Just found this definition. For $$M=\left[\begin{array}{cc}{U_{1}} & {U_{2}}\end{array}\right]\left[\begin{array}{cc}{\Sigma_{1}} & {0} \\ {0} & {0}\end{array}\right]\left[\begin{array}{c}{V_{1}^{\prime}} \\ {V_{2}^{\prime}}\end{array}\right]$$ $$\text { Nullspace: } \mathcal{N}=\operatorname{span}\left[V_{2}\right]$$ $$\operatorname{image}(M)=\operatorname{span}\left[U_{1}\right]$$

If I got this right the span is the spanning set of a matrix. I see how I get it despite a lack of intuition about it. What I don't get is why the are writing those indices. Isn't $$V_2$$ a vector? Whats the span of a vector and why the second one? The same for $$U$$. I already got this answer

the vectors $$u_1,\dots,u_r$$ will form an orthonormal basis for the image of $$P$$ and the vectors $$v_{r+1},\dots,v_{n}$$ will form a basis for the kernel of $$P$$.

from @Omnomnomnom. They are somehow similar but not identical. Omnomnomnom talks about a set of vectors whilst this definition talks about the span of a vector?

As you said, the rank of the matrix will simply be the number of non-zero singular values. Let $$r$$ denote this rank.
Given an SVD $$P = U\Sigma V^T$$ with $$u_i,v_i$$ denoting the $$i$$th column of $$U,V$$ respectively, the vectors $$u_1,\dots,u_r$$ will form an orthonormal basis for the image of $$P$$ and the vectors $$v_{r+1},\dots,v_{n}$$ will form a basis for the kernel of $$P$$.
• @Mr.Sh4nnon Here's one explanation: the only non-zero outputs of $P$ must be linear combinations of the directions $u_i$ that correspond to non-zero stretch-factors $\sigma_i$. Sep 4, 2019 at 15:19
• In the context you found, $V_1,V_2$ are block matrices. The columns in $V_2$ are (in my notation) $v_{r+1},\dots,v_n$, so the answers do really say the same thing. Sep 7, 2019 at 18:52
• @Mr.Sh4nnon keep in mind that $V'$ is another way of writing $V^T$ (transposed) in this context, so yes: $v_{r+1},\dots,v_n'$ are rows, and $v_{r+1},\dots,v_n$ are columns. Sep 8, 2019 at 15:55