Find a Matrix $P$ such that $PA=\mathbf 0$ (zero matrix)

Question

Can P be considered as the null space of the matrix A or is it a projection onto the null-space of the matrix. I am a little confused with the basic concepts behind linear algebra when it comes to sub-spaces so kindly be a little elaborate. (A and P are both matrices)

Answer 1

Let $A$ be an $(m\times n)$-matrix and $\mathrm{span}(A)$ the linear hull of its columns. This is some subspace $V$ of $\Bbb R^m$. Choose some orthogonal subspace $W\subseteq V^\top$, i.e. a subspace of $\Bbb R^m$ spanned by vectors $w_i$ with

$$w_i^\top v=0,\quad\text{for all $v\in V$}.$$

Build your matrix $P$ as an $(k\times m)$-matrix with the rows $w_i$. Then $PA=\mathbf 0$. And yes, as you said this is exactly a projection onto the null-space $\ker(A)=V^\top$ of $A$. $P$ itself is a matrix and not a space, so not the null-space itself, but it rows span a subspace $W$ of the null-space.

If $V=\Bbb R^m$, then your only choice for $W$ is the zero space $\{0\}$, i.e. $P=\mathbf 0$. This is always a solution, but for some matrices there might be more ways do define $P$.

As your question was a bit confusing on whether your $A$ is a matrix or a vector: if $A$ should be a vector, then just choose $n=1$ in the discussion above. This means you choose the $w_i$ just in a way to be orthogonal to your vector $A$. In general, one does not talk about zero-spaces for vectors because vectors are not maps (but can be associated with such).

Answer 2

Maybe it is easier to consider the induced linear map $L_Av:= Av$ and $L_Pv=Pv$. Since $PA=0$ you get $L_PL_A\equiv0$. If $v\in ker L_A$, then you already get $L_PL_Av=L_P0=0$. But if $w\in image(L_A)\setminus\{0\}$, then there exists $v$ such that $w=L_Av\neq 0$ and you would like to have $0=L_PL_Av=L_Pw$. This means that you need $ker L_P\supset image(L_A)$.

In order to get P, you have to compute the image of $L_A$ and define $P$ such that the kernel of $L_P$ contains the image of $L_A$. Obviously $P=0$ will do it. But if $A$ is not invertable then $L_P$ as the orthogonal projection on $(image(L_A))^\bot\neq\{0\}$ will do it and $L_p\not\equiv 0$ as $P\neq 0$.

Answer 3

When we talk about null-space it's something that relates to a mapping. When talking matrices and vectors it's the former that normally correspond to mapping.

Also the mapping notation is normally that one writes the mapping first and the vector last. For example we write $Ax$ to denote what the mapping $A$ of $x$ is. This convention is also used when it comes to matrices.

This means that the nullspace of a matrix $M$ is all vectors $x$ such that $Mx=0$.

Your notation seem to be somewhat backward or just confused (as you mention the null-space of a vector).

On the other hand if we have a matrix $P$ such that $AP=0$ we at least have that $P$ will map into the null-space of $A$, but it will not automatically make $P$ be a projection. For a matrix or mapping to be a projection we also need $P^2=P$ to hold which doesn't happen automatically. Take for example consider $A=0$ and $P$ is a rotation matrix (which means that $P^2\ne P$ in general).