Filetype PDF | Posted on 25 Jan 2023 | 2 years ago
Authentication
digital resources
160x Filetype PDF File size 2.77 MB Source: math.northwestern.edu
Math 290-1: Linear Algebra & Multivariable Calculus
Northwestern University, Lecture Notes
Written by Santiago Can˜ez
These are notes which provide a basic summary of each lecture for Math 290-1, the first quarter
of “MENU: Linear Algebra & Multivariable Calculus”, taught by the author at Northwestern
University. The book used as a reference is the 5th edition of Linear Algebra with Applications by
Bretscher. Watch out for typos! Comments and suggestions are welcome.
Contents
Lecture 1: Introduction to Linear Systems 2
Lecture 2: Gauss-Jordan Elimination 3
Lecture 3: Solutions of Linear Systems 7
Lecture 4: More on Solutions of Systems and Vectors 11
Lecture 5: Linear Transformations 14
Lecture 6: Geometric Transformations 19
Lecture 7: Matrix Multiplication 22
Lecture 8: Invertibility and Inverses 26
Lecture 9: More on Inverses, the Amazingly Awesome Theorem 30
Lecture 10: Images and Kernels 33
Lecture 11: Subspaces of Rn 38
Lecture 12: Linear Dependence/Independence and Bases 40
Lecture 13: Bases and Dimension 45
Lecture 14: Coordinates Relative to a Basis 49
Lecture 15: More on Coordinates 55
Lecture 16: Determinants 60
Lecture 17: Properties of Determinants 65
Lecture 18: Geometric Interpretation of Determinants 70
Lecture 19: Eigenvalues 75
Lecture 20: Eigenvectors 79
Lecture 21: Applications of Eigenvectors 85
Lecture 22: Diagonalization 91
Lecture 23: More on Diagonalization 96
Lecture 24: Complex Eigenvalues 102
Lecture 1: Introduction to Linear Systems
TodayIgaveabriefintroduction to some concepts we’ll be looking at this quarter, such as matrices,
eigenvalues, and eigenvectors. I mentioned one or two ways in which such concepts show up in other
areas.
Example 1. The system of linear equations (also known as a linear system):
x+2y=0
−3x−2y=8
has precisely one solution: x = −4,y = 2. Geometrically, both of these equations describe lines in
the xy-plane and the existence of only one solution means that these two lines intersect in exactly
one point.
Example 2. The system of linear equations:
x+2y=0
−3x−6y=−3
has no solutions. Geometrically, this happens because the corresponding lines are parallel and don’t
intersect.
Example 3. The system of equations:
x+2y=0
−3x−6y=0
has infinitely many solutions, meaning that there are infinitely many pairs of numbers (x,y) which
satisfy both equations simultaneously. Geometrically, these two equations describe the same line
and so intersect everywhere.
Important. The same phenomena regarding number of solutions is true in any number of dimen-
sions. In other words, any system of linear equations no matter how many variables or equations
are involved will have exactly one solution, no solution, or infinitely many solutions.
Example 4. Consider the system:
x+ 2y+3z=0
−3x− 2y−8z=8
2x+12y+ z=2
Geometrically, each of these equations describe planes in 3-dimensional space (we’ll talk about
planes a lot more when we get to multivariable calculus) and by finding the solution(s) of this
system we are determining where these three planes intersect. We solve the system using what are
called “row operations”, and we’ll describe this method in detail next time.
For now, note that multiplying the first equation by 3 gives 3x + 6y + 9z = 0, and adding this
entire equation to the second one gives 4y + z = 8. The point is that this resulting equation no
longer has an x in it, so we’ve “eliminated” a variable. Similarly, multiplying the first equation
by −2 gives −2x − 4y − 6z = 0 and adding this to the third gives 8y − 5z = 2, and again we’ve
2
eliminated x. Now consider the system keeping the first equation the same but replacing the second
and third with the new ones obtained:
x+2y+3z=0
4y + z = 8
8y −5z = 2
The point is that this new system has precisely the same solutions as the original one! In other
words, “row operations” do change the actual equations involved but do not change the set of
solutions.
We can keep going. Now we move down to the 4y terms and decide we want to get rid of the
8y below it. We multiply the second equation by −2 and add the result to the third equation to
give −7z = −14. Thus we get the new system
x+2y+3z=0
4y + z = 8
−7z=−14
Nowwe’re in business: the third equation tells us that z = 2, substituting this into the second and
solving for y gives y = 3/2, and finally substituting these two values into the first equation and
solving for x gives x = −9. Thus this system has only solution:
x=−9,y=3/2,z=2.
Again, since this method does not change the solutions of the various systems of equations we use,
this is also the only solution of our original system.
Lecture 2: Gauss-Jordan Elimination
Today we started talking about Gauss-Jordan Elimination, which gives us a systematic way of
solving systems of linear equations. This technique is going to be the most useful computational
tool we’ll have the entire quarter, and it will be very beneficial to get to the point were you can
carry it out fairly quickly and without errors. Practice makes perfect! We’ll continue with examples
on Monday.
Warm-Up 1. Solve the system of equations:
2x+3y+z=0
x− y+z=2
Weuse the technique of “eliminating” variables. We first multiply the second row by −2 and add
the first row to it, giving 5y − z = −4. So now we have the system
2x+3y+z=0
5y −z = −4
Now there are multiple ways we could proceed. First, we could add these two equations together
and use the result to replace the first equation, giving:
2x+8y =0
5y −z = −4
3
Compared to our original set of equations, these are simpler to work with. The question now
is: what do we do next? Do we keep trying to eliminate variables, or move on to trying to find
the solution(s)? Note that any further manipulations we do cannot possibly eliminate any more
variables, since such operations will introduce a variable we’ve already eliminated into one of the
equations. We’ll see later how we can precisely tell that this is the best we can do. So, let’s move
towards finding solutions.
For now, we actually go back to equations we had after our first manipulations, namely:
2x+3y+z=0
5y −z = −4
Wecould instead try to eliminate the y term in the first equation instead of the z term as we did.
This illustrates a general point: there are often multiple ways of solving these systems, and it would
be good if we had a systematic way of doing so. This is what Gauss-Jordan elimination will do for
us. Here, let’s just stick with the above equations.
Wewill express the values of x and y in terms of z. The second equation gives
y = z −4.
5
Plugging this in for y in the first equation and solving for x gives:
−3y−z −3!z−4"−z 12−8z
x= = 5 = .
2 2 10
These equations we’ve derived imply that our system in fact has infinitely many solutions: for any
value we assign to z, setting x equal to 12−8z and y equal to z−4 gives a triple of numbers (x,y,z)
10 5
which form a solution of the original equation. Since z is “free” to take on any value, we call it a
“free” variable. Thus we can express the solution of our system as
x=12−8z, y= z−4, z free.
10 5
Warm-Up2. Findthepolynomialfunctionoftheformf(x) = a+bx+cx2 satisfyingthecondition
that its graph passes through (1,1) and (2,0) and such that #2f(x)dx = −1.
1
The point of this problem is understanding what this has to do with linear algebra, and the
realization that systems of linear equations show up in many places. In particular, this problem boils
down to solving a system of three equations in terms of the three unknown “variables” a,b, and c.
The condition that the graph of f(x) pass through (1,1) means that f(1) should equal 1 and the
condition that the graph pass through (2,0) means that f(2) should equal 0. Writing out what
this means, we get:
f(1) = 1 means a+b+c = 1
and
f(2) = 0 means a+2b+4c = 0.
Finally, since $ % &'
2 bx2 cx3 '2 3 7
(a+bx+cx2)dx= ax+ + ' =a+ b+ c,
#1 2 3 '1 2 3
the condition that 2 f(x)dx = −1 gives
1
a+3b+7c=−1.
2 3
4
no reviews yet
Please Login to review.
