§1 – Row Space, Column Space, and Kernel
Starting with a matrix . . .
Given an m × n matrix A , we considered three important subspaces. The row space row( A ) is the span of
the rows of A , the column space col( A ) is the span of the columns of A , and the kernel ker( A ) is the set of
all ~x ∈ Rn that satisfy the matrix equation A~x = ~0. In other words, it is the general solution to the linear
system with augmented matrix [ A | ~0 ]. In your textbook, kernels are called nullspaces. It is often required
to determine bases for these vector spaces. Do so using the algorithms introduced in class. Please don’t
forget to write your final answers like “A basis for . . . is { . . . }.” — At that point, don’t use the symbol “span”
and don’t use parameters like “ s · . . . + t · . . .”.
W15–T4 # 5 ? F14–T5 # 4 F15–T4 # 4
Extend the basis you found in W15–T4 # 5a to a basis for R . ?
Extend the basis you found in W15–T4 # 5b to a basis for R . ?
4
3
H INT: In W15–T4 # 5c and in F14–T5 # 4c you are asked to determine the dimension of the following
subspace: { A~x | ~x ∈ Rn }. This subspace is one of row( A ), col( A ), ker( A ). Recall the different ways of
interpreting the product of a matrix A and a column vector ~x to identify which one it is.
Dimensions
In light of the different algorithms, assume given an 11 × 7 matrix A with rank( A ) = 5. Please fill in the
boxes in the following two tables:
row( A ) ⊆ R col( A ) ⊆ R dim(row( A )) = dim(col( A )) = ker( A ) ⊆ R ker( A T ) ⊆ R dim(ker( A )) = dim(ker( A T )) = H INT: When thinking about ker( A T ), keep in mind that the transposed matrix A T is obtained from A
by turning rows into columns. In particular, we know that row( A ) is precisely the same as col( A T ). This
has the nice consequence that rank( A ) = dim(row( A )) = dim(col( A T )) = rank( A T ). You may now apply
the rank nullity theorem to determine dim(ker( A T )).
Starting with a spanning set . . .
Let W = span{ (3, 1, −2), (−6, −2, 4), (1, 4, 0), (7, 6, −4) }.
Find a basis for W that is a subset of { (3, 1, −2), (−6, −2, 4), (1, 4, 0), (7, 6, −4) }. ?
Give a geometric description of W . ?
H INT: A complete geometric description consists of four pieces of information: Is the subspace a point,
or a line, or a plane, . . . ? Is the surrounding space R2 , or R3 , . . . ? Through which point does the subspace
go? What are direction vectors or normal vectors? So, a complete answer to a problem of that kind could
look like this: “W is a line in R2 going through the origin with direction vector (1, 2).”
Bonus: Finding a matrix with a particular kernel
We could also go the other way and, given a subspace, try to find a matrix with that subspace as a kernel.
This is something that we have not done in class but that you can do based on what we did. So, let W be
the subspace of R4 with basis { (1, 2, 0, 1), (0, 0, 1, 3) }.
Find a 2 × 4 matrix A with ker( A) = W .
H INT: Recall that the rows of a matrix are always orthogonal to the vectors in the kernel. So, looking for
possible rows, our first step is to find all vectors (a, b, c, d ) ∈ R4 that are both orthogonal to (1, 2, 0, 1) and
orthogonal to (0, 0, 1, 3). This can be done by solving a homogeneous linear system with two equations.
The general solution of this system contains all possible candidates for the rows of A . Because of the
rank nullity theorem, we know that rank( A ) must be 2. Let us therefore pick two linearly independent
vectors from our candidates and write them into the rows of the matrix A . Then, by construction,
ker( A ) contains both (1, 2, 0, 1) and (0, 0, 1, 3) and, by the rank nullity theorem, dim(ker( A )) = 2. So,
{ (1, 2, 0, 1), (0, 0, 1, 3) } is a linearly independent set with two vectors in a two dimensional space, and
therefore a basis for ker A . In other words, ker( A ) = W .
§2 – Arithemtic Porperties of Matrix Multiplication
and the Invertible Matrix Theorem
Arithmetic properties
Recall the arithmetic properties of matrix multiplication. In particular, they say that we are only allowed
to multiply certain matrices, the number of columns of the first one has to be equal to the number of rows
of the second one. Moreover, matrix multiplication is not commutative. Imagine a true-false question with
the following statement: “For any two n× n matrices A and B, the binomial formula ( A +B)2 = A 2 +2 AB +B2
holds.”
Justify why the statement is false? ?
What is wrong with the following solutions? ?
Invertible matrices
An n × n matrix A is called invertible if there is an n × n matrix B such that AB = I and BA = I . In fact,
we just need to check that AB = I . Let us prove this carefully:
Show that if there is an n × n matrix B such that AB = I , then A is invertible.
H INT: In class, we assumed that A ∼ I and showed there is an n × n matrix B such that both AB = I
and BA = I hold. Now, we don’t know whether A ∼ I but we may assume that AB = I . Therefore, we
know that each column of I is a linear combination of the columns of A . So, the column space col( A )
contains a linearly independent set of n vectors. So, dim(col( A )) = n. So, A is invertible. Your job is to
fill in the details and justify each step!
Now, let A and B be any two n × n matrices.
Show that if AB is invertible, then A is invertible. ?
H INT: Use the above result. It suffices to find a matrix M such that AM = I . Try to construct such a
matrix M , maybe as a product whose factors are matrices like A , B, or ( AB)−1 .
F14–T5 # 6ab ? F14–T5 # 2 F14–T5 # 3
§3 – Matrices and Linear Systems
VSF Problem # 14.7jklm ?