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Course Notes for Linear Algebra

Remkes Kooistra

Section 7.3 Kernels and Images

Subsection 7.3.1 Row and Column Spaces

Definition 7.3.1.

Let \(M\) be an \(m \times n\) matrix. The rowspace of the matrix is the span of the rows of the matrix (considered as vectors in \(\RR^n\)). It is a subspace of \(\RR^n\text{.}\) The columnspace of the matrix is the span of the columns of the matrix, considering those columns as vectors in \(\RR^m\text{.}\) It is a subspace of \(\RR^m\text{.}\)

Subsection 7.3.2 Images

In Section 7.2, I defined the image of a matrix or transformation to be the whole collection of outputs. (This is what I call the range in single variable functions.) There is a nice proposition which tells me how to calculate images for linear transformations.

Proof.

The image is the set of all outputs of \(M\text{,}\) acting on the whole space \(\RR^n\text{.}\) I can think of \(\RR^n\) as the span of its standard basis \(e_1, e_2, \ldots e_n\text{.}\) The image, then, is the span of \(Me_1, Me_2, \ldots Me_n\text{.}\) These vectors are precisely the columns of the matrix, so their span is the columnspace.

Example 7.3.4.

Here is a \(3 \times 3\) matrix acting on the standard basis of \(\RR^3\text{.}\) I can demonstrate how the images of the basis vectors are exactly the columns of the matrix.
\begin{align*} \begin{pmatrix} -2 \amp -2 \amp 0 \\ 7 \amp -3 \amp 2 \\ -4 \amp -1 \amp 2 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 7 \\ -4 \end{pmatrix}\\ \begin{pmatrix} -2 \amp -2 \amp 0 \\ 7 \amp -3 \amp 2 \\ -4 \amp -1 \amp 2 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -3 \\ -1 \end{pmatrix}\\ \begin{pmatrix} -2 \amp -2 \amp 0 \\ 7 \amp -3 \amp 2 \\ -4 \amp -1 \amp 2 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} \end{align*}

Subsection 7.3.3 Kernels

Definition 7.3.5.

Let \(M\) be an \(m \times n\) matrix representing a transformation \(\RR^n \rightarrow \RR^m\text{.}\) For any \(u \in \RR^m\text{,}\) the preimage of \(u\) is the set of all vectors in \(\RR^m\) which are mapped to \(u\) by \(M\text{.}\) It is writtem \(M^{-1}\{u\}\text{.}\) The kernel or nullspace of \(M\) is all vectors in \(\RR^n\) which are sent to the zero vector under the transformation associated to \(M\text{.}\) Combining the two terms, the kernel is the preimage of the zero vector.

Proof.

The preimage is all vectors \(v\) in \(\RR^n\) that are sent to \(u\text{.}\) If I write \(v\) in coordinates \(x_1, \ldots, x_n\text{,}\) and apply the matrix action, I get a system of \(m\) equations with the entries of \(u\) as the constants. The matrix encoding of this system is precisely the extended matrix \((M|u)\text{.}\)
I can look at \(\RR^2\) to build intuition. Conider a general matrix action in two variables.
\begin{equation*} \begin{pmatrix} a \amp b \\ c \amp d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} \end{equation*}
If I write this out in terms of coordinates, we get a system.
\begin{align*} ax + by \amp = x_0\\ cx + dy \amp = y_0 \end{align*}
This is the system associated to the extended matrix with \(\begin{pmatrix} x_0 \\ y_0 \end{pmatrix}\) as constants.

Subsection 7.3.4 A Central Connection Idea

At this point in the course, I have talked about loci, solved linear systems, and now defined the preimage of a vector under a transformation. Loci are very geometric, systems are very algebraic, and preimages are again geometric but from a different perspective (transformations instead of static objects). Despite their different starting points, these three are essentially all talking about the same thing. The connection between all three is a major conceptual goal in this course; if I can understand how these three novel and abstract definitions fit together, that points me to a deep and thorough understanding of the three concepts. Let me try to describe this connection.
Let me start with a linear system. This is a set of linear equations, implicitly asking what values of the variables (if any) solve all the equations simultaneously. To move to loci, all I have to do is think geometrically. A locus is the set of points that satisfy some equations (linear equations in this course). So a locus is just the solution of a system, seen geometrically. A locus is a solution space. I made this connection before when I calculated the dimension of loci: the dimension was the number of free parameters in the solution of the associated system.
So what about pre-images? Conisider a \(m \times n\) matrix \(M\text{,}\) representing a transformation \(\RR^m \rightarrow \RR^n\text{.}\) For a vector \(u\) in \(\RR^n\text{,}\) I can ask for the pre-image \(M^{-1}\{ u\}\text{.}\) This is all the vectors in \(\RR^m\) which the matrix sends to \(u\) (this could possibly be empty if none of the vectors are sent to \(u\)). I can express this as a matrix equation: \(v\) is in the preimage only if
\begin{equation*} Mv = u\text{.} \end{equation*}
But then I can look at the coordinates of this equation. Each coordinate gives me a linear equation, so I get a system of \(m\) linear equations (because \(v \in \RR^m\text{,}\) so it has \(m\) coordinates). The constants of this linear system are the components of \(u\text{.}\) The matrix of this linear system is the extended matrix \((M|u)\text{.}\) The solution to this linear system can be interpreted as a locus. This locus (line, plane, etc) is precisely all the points that \(M\) sends to \(u\text{.}\) In this way, solutions spaces to systems, loci, and preimages are all the same things and are all found by the matrix \((M|u)\text{.}\)
Specifically, the kernel is the linear space of all vectors \(v\) with \(Mv = 0\text{.}\) This is a solution space for the linear system with matrix \((M|0)\text{,}\) where the column after the dividing line is all zeroes. To calculate a kernel, I have to solve a system where the constants are all zero. The matrix of that system is just \(M\text{,}\) the matrix of the transformation.
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