Deducing properties of a transformation from its matrix Give the rank of the matrix
$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
Is the corresponding linear mapping injective, surjective,
bijective?
Answer: the rank is three. Thus, the corresponding linear mapping is
neither injective, nor surjective or bijective.
It is clear that the matrix has a rank of 3, since there are only three linearly independent columns in it. However, where do we get the properties of linear mapping and where is the mapping defined anyway? (don't see any corresponding notations) I'm rather new to this, so any readings are of great value. Thanks in advance.
 A: An $n\times m$ matrix is interpreted as an $F$ linear transformation from $F^n$ to $F^m$ by multiplying row vectors from $F^n$ on their right side by this matrix. The result is a row vector in $F^m$. (Alternatively this can all be done with column vectors mutliplied on the left by this matrix, and then the map would be from $F^m$ to $F^n$.)
You should be able to see that it is indeed possible to find a vector which is sent to zero by this matrix, proving it is not injective.
If you know the so-called "rank-nullity theorem," it should connect this quite well with the injective and surjective conditions. It tells you how the rank of the matrix (which is the dimension of the image of the transformation) is connected with the dimension of the kernel and the dimension of the codomain (the space the function is going to). 
A: I guess the mapping is defined from/to $\mathbb{R}^3 \to \mathbb{R}^3$ if this is from a first course in linear algebra. I don't know how far you have come in your linear algebra course, but there is a theorem which says that a $n \times n$ matrix gives an injective linear map if and only if it gives a surjective map if and only if the matrix has $n$ linearly independent columns if and only if the matrix has a determinant different from $0$. This theorem doesn't necessarily hold if the matrix is not square i.e it is not an $n \times n$ matrix. You also have the famous rank theorem which states that for a linear operator $T$ between finite dimensional vector spaces you have that $rank(T) + dim \; ker(T) = dim \text{(dimension of the space the operator is defined on)}$.
A: Here's one way to look at it.
Consider a differentiable function $f$ mapping $\Bbb R^n$ to $\Bbb R^m$.
As with single-variable functions, the derivative of $f$ is a linearization of $f$ around some point $\bf{x_0}$. The derivative of $f$ in this case is a matrix. So we write
$$f(\mathbf{x}+\mathbf{h}) = f(\mathbf{x})+f'(\mathbf{x})\mathbf{h} + o(\mathbf{h}^2)$$
One consequence of the locally linear nature of differentiability is that there is a neighborhood around $\mathbf{x_0}$ such that $f'$ is invertible. This is the core idea behind the Inverse Function Theorem.
So, if the function has a local inverse, then its derivative is invertible in this neighborhood. An invertible function must be bijective -- injective and onto -- and hence so must this linear mapping.
