Is every linear transform that takes a non-zero vector to zero composed of a projection? I'm reviewing some linear algebra, and this question came up.
If $T$ is a transformation matrix in $\mathbb{R}^{n\times n}$, and $v$ is some non-zero vector in $\mathbb{R}^n$, and $vT = 0$, then the are a whole bunch of other vectors that are also taken to zero; at least all vectors $av$ with $a \in \mathbb{R}$). It seems to me that this can only happen if at least one dimension is "squashed". (So if you are in $\mathbb{R}^3$, for example, and one vector goes to 0, then all vectors go to a plane.) 
Is this in fact true? 
One type of matrix that does this is projection matrices (where $T^2 = T$), and of course products like $PT$, where $P$ is a projection matrix and $T$ is any matrix. 
So my main question is, if every transformation that takes some non-zero vector to 0 is of this form: $PT$ or $TP$, where $P$ is a projection matrix, and $T$ is a linear transform that only takes 0 (and no other vector) to 0.
(My background is in engineering / programming, and I have only limited knowledge of more abstract concepts.)
 A: Very good question!
About whether dimensions get "squashed", yes, it is true. The precise formulation is the rank nullity theorem, which says that the dimension of the image (which is two if the image is a plane, for example) plus the dimension of the kernel (all those vectors squashed to zero) is equal to the dimension of the space you started with. So the more vectors you send to zero, the smaller an image you will have. 
In response to your question about projections: in a sense it is true, but in a slightly different way from the way that (I think) you mean. Suppose $T:V \to W$ is a linear transformation. Let's think about the possibility of $T$ being a projection followed by an injective (i.e. kernel-free) linear map: a first guess might be to try to project onto those vectors that are not sent to zero by $T$. But this is not even a subspace! (Since any subspace contains zero, and zero always goes to zero.) So the vectors not sent to zero can't be the image of a linear map. But there is another candidate: instead use a quotient space of $V$, namely $V/\operatorname{ker}T$. Then $T$ is a projection followed by an injective map from $V/\operatorname{ker}T$ to $W$.
As far as $T$ being an injective function followed by a projection, we have less chance of this, because what if the image of $T$ is all of $W$, but $T$ is still not injective? Then the rank nullity theorem tells us that there is no way to create an injective function from $V$ to $W$ (if $V$ and $W$ are both finite-dimensional), so there are no candidates for an injective function to start with. 
A: The way you can think about it is like this. If you are in $\mathbb{R}^3$ and you are given some vector $v\in\mathbb{R}^3$ such that $Tv=0$, then take the basis $B=\{b_1,b_2,v\}$ for $\mathbb{R}^3$. Since any vector $y\in\mathbb{R}^3$ can be written as $y=c_1b_1+c_2b_2+c_3v$ for some $c_1,c_2,c_3\in\mathbb{R}$ and since linear transformations act linearly, at the least, $T$ eliminates the vector $v$ from every $y\in\mathbb{R}^3$ thus leaving at a maximum a $2$-dimensional space. This new space will be spanned by $Tb_1$ and $Tb_2$. 
