§1 – Matrix Multiplication
If A ∈ Mmn (“ A is an m × n matrix”) and B ∈ Mnk (“B is an n × k matrix”), then we may compute their
product AB ∈ Mmk . By now, all of you should know that the default way to do so is to compute the dot
product of the i -th row of A and the j -th column of B, which gives the ij -th entry of AB. An example for
this row-times-column procedure is:
http://mysite.science.uottawa.ca/jcuno/teaching/winter17/mat1341/lecture-15.pdf
Now, it’s your turn. Decide if the following products are defined and, if so, compute the result:
·
1
3
5
7
£3
¸
2
1
1
¤
4
· 1
0
·
·

2
5

0
1  = ...
3
¸
= ...
·
3
1
·
¸
·
3
1
£
¸ ·
¸
2
·
= ...
5
2
5
¤
= ...
Instead of using the row-times-column procedure we could also have obtained the columns of AB as linear
combinations of the columns of A and we could have obtained the rows of AB as linear combinations of the
rows of B. In the first of the four examples above, the first column of B has the entries 4, 0, 1. Therefore,
the first column of AB must be “4 · first column of A + 1 · second column of A + 0 · third column of A ”. This
kind of issue is often addressed in tests:
W15–T4 # 1 ? F15–T3 # 2 ?
We have seen in class that matrix multiplication is not commutative. We may thus fix a particular matrix
S ∈ M22 and ask which matrices A ∈ M22 commute with S .
F15–T4 # 3 ?
H INT: Consider an arbitrary 2 × 2 matrix:
·
A=
a
c
b
d
¸
Evaluate both S A and AS . We are interested in the set W = { A ∈ M22 | S A = AS }. So, the two products
S A and AS must be equal, and therefore all their entries must be equal. This yields a set of conditions
on a, b, c, d ∈ R that need to be satisfied. Once you have found these conditions, you may use them to
write the set W as a span.
§2 – Kernel and the Rank Nullity Theorem
In class, we have computed the kernel of a matrix. For example on slides 25 – 56 of the last lecture:
http://mysite.science.uottawa.ca/jcuno/teaching/winter17/mat1341/lecture-16.pdf
You should be able to do this on your own:
W15–T4 # 5d ?
True or False? — If A is a matrix, then every row of A
is orthogonal to every vector in ker( A ). ?
An important consequence of our studies of the kernel was the rank nullity theorem, which said that, for
every matrix A , the equation rank( A ) = # columns of A − dim(ker( A )) holds. So, when we are just interested
in the dimension of the kernel, it suffices to know the number of columns and the rank (Example 1). On
the other hand, when we are interested in the rank, then it suffices to know the number of columns and
the dimension of the kernel (Example 2).
F15–T3 # 6a ? F14–T4 # 8
H INT: For F14–T4 # 8, there are several solutions. The following strategy is very elegant. You know
that the columns of the matrix are linearly independent. In other words, x1 = x2 = 0 is the only solution
to the vector equation:
·
¸
·
¸ ·
¸
a
c
0
x1
+ x2
=
b
d
0
We can translate between vector and matrix equations. So we also know that x1 = x2 = 0 is the only
solution to the matrix equation:
·
¸·
¸ ·
¸
a b
x1
0
=
c d
x2
0
This tells you something about the kernel! Then, use the rank nullity theorem.
§3 – Matrix Inverses
We have also talked about invertability of n × n matrices and learnt an algorithm to compute the inverse
using superaugmented matrices.
F15–T4 # 2 ? W15–T4 # 2
It turned out that the existence of an inverse simplified a lot of things. For example, in order to solve the
matrix equation A~x = ~
b, we just have to multiply from the left by A −1 and obtain ~x = A −1~
b. Try to use this
kind of tricks to solve:
F15–T4 # 6a(i) ? F14–Final # 15a(i) ?
Then, towards the end of the last lecture, we saw a couple of statements that were all equivalent to “ A
is invertible”. That means either all of them are true or all of them are false. So, knowing that A is
invertible, we know that the columns of A are linearly independent and a lot of other things. This should
be practiced:
W15–T4 # 6b ? F15–T4 # 6b
§A – Matrix Powers (Bonus)
Given an n × n matrix A , we may multiply the matrix by itself and thus compute matrix powers such as
A 2 = A A and A 3 = A A A . The matrix A is called nilpotent if some matrix power A k is the zero matrix.
Convince yourself that the following matrix A is nilpotent:

0
A= 0
0
1
0
0
F15–T3 # 6c

0
1 
0
Now, consider the following matrix:
·
Q=
1
1
1
0
¸
Compute its matrix powers Q , Q , . . . , Q .
2
3
5
? Do you recognize a pattern?
H INT: Recall that the Fibonacci numbers. This is the sequence of numbers that is recursively defined
by F0 := 0, F1 := 1 and, for all n > 2, F n := F n−1 + F n−2 . Write the first Fibonacci numbers down. When
you are looking for a pattern, try to express the matrix powers Q 2 , Q 3 , . . . , Q 5 in terms of Fibonacci
numbers. If you want to learn more about Fibonacci numbers, watch this video:
https://www.youtube.com/watch?v=SjSHVDfXHQ4 ( 6:24 )
§B – Approaching the Final Exam (Bonus)
Our class is slowly going towards the end. And it is time to start familiarizing yourself with the previous
final exams. Of course, there is still a lot of material to be covered. But you may already answer a lot of
problems:
F15–Final # 1 # 2 # 4 # 5 # 6 # 7 # 11 # 12ab # 15a(ii)
W15–Final # 1 # 2 # 3 # 4 # 5 # 6 # 7 # 11 # 15abc
I have written detailed solutions for you:
Details F15–Final
Details W15–Final