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LINEAR ALGEBRA IN A NUTSHELL
One question always comes on the first day of class. “Do I have to know linear
algebra?” My reply gets shorter every year: “You soon will.” This section brings
together many important points in the theory. It serves as a quick primer, not an
official part of the applied mathematics course (like Chapter 1 and 2).
This summary begins with two lists that use most of the key words of linear
algebra. The first list applies to invertible matrices. That property is described in 14
different ways. The second list shows the contrast, when A is singular (not invertible).
There are more ways to test invertibility of an n by n matrix than I expected.
Nonsingular Singular
A is invertible A is not invertible
The columns are independent The columns are dependent
The rows are independent The rows are dependent
The determinant is not zero The determinant is zero
Ax = 0 has one solution x = 0 Ax = 0 has infinitely many solutions
Ax = b has one solution x = A
− 1 b Ax = b has no solution or infinitely many
A has n (nonzero) pivots A has r < n pivots
A has full rank A has rank r < n
The reduced row echelon form is R = I R has at least one zero row
The column space is all of R
n The column space has dimension r < n
The row space is all of R
n The row space has dimension r < n
All eigenvalues are nonzero Zero is an eigenvalue of A
A
T A is symmetric positive definite A
T A is only semidefinite
A has n (positive) singular values A has r < n singular values
Now we take a deeper look at linear equations, without proving every statement
we make. The goal is to discover what Ax = b really means. One reference is my
textbook Introduction to Linear Algebra, published by Wellesley-Cambridge Press.
That book has a much more careful development with many examples (you could look
at the course page, with videos of the lectures, on ocw.mit.edu or web.mit.edu/18.06).
The key is to think of every multiplication Ax, a matrix A times a vector x, as a
combination of the columns of A:
Matrix Multiplication by Columns
[
] [
C
D
]
= C
[
]
+ D
[
]
= combination of columns.
Multiplying by rows, the first component C + 2D comes from 1 and 2 in the first row
of A. But I strongly recommend to think of Ax a column at a time. Notice how
[ ] [ ] [ ] [ ]
x = (1, 0) and x = (0, 1) will pick out single columns of A:
= first column = last column.
Suppose A is an m by n matrix. Then Ax = 0 has at least one solution, the all-zeros
vector x = 0. There are certainly other solutions in case n > m (more unknowns
than equations). Even if m = n, there might be nonzero solutions to Ax = 0; then
A is square but not invertible. It is the number r of independent rows and columns
that counts. That number r is the rank of A (r ≤ m and r ≤ n).
The nullspace of A is the set of all solutions x to Ax = 0. This nullspace
N (A) contains only x = 0 when the columns of A are independent. In
that case the matrix A has full column rank r = n: independent columns.
For our 2 by 2 example, the combination with C = 2 and D = − 1 produces the zero
vector. Thus x = (2, −1) is in the nullspace, with Ax = 0. The columns (1, 3) and
(2, 6) are “linearly dependent.” One column is a multiple of the other column. The
rank is r = 1. The matrix A has a whole line of vectors cx = c(2, −1) in its nullspace:
[ ] [ ] [ ] [ ] [ ] [ ]
Nullspace
is a line
and also
2 c
−c
If Ax = 0 and Ay = 0, then every combination cx + dy is in the nullspace. Always
Ax = 0 asks for a combination of the columns of A that produces the zero vector:
x in nullspace x 1 (column 1) + · · · + xn (column n)= zero vector
When those columns are independent, the only way to produce Ax = 0 is with x 1
x 2
= 0,.. ., x n
= 0. Then x = (0,... , 0) is the only vector in the nullspace of A.
Often this will be our requirement (independent columns) for a good matrix A. In
that case, A
T A also has independent columns. The square n by n matrix A
T A is then
invertible and symmetric and positive definite. If A is good then A
T A is even better.
I will extend this review (still optional) to the geometry of Ax = b.
Column Space and Solutions to Linear Equations
Ax = b asks for a linear combination of the columns that equals b. In our
2 by 2 example, the columns go in the same direction! Then b does too:
[ ] [ ] [ ]
Column space Ax =
C
D
is always on the line through
We can only solve Ax = b when the vector b is on that line. For b = (1, 4) there is no
solution, it is off the line. For b = (5, 15) there are many solutions (5 times column 1
gives b, and this b is on the line). The big step is to look at a space of vectors:
[ ] [ ]
[ ] [ ] [ ] [ ] [ ]
- Ax = b has infinitely many solutions in Figure A1. The shortest x always lies
in the “row space” of A. That particular solution (1, 2) is found by the pseudo-
inverse pinv (A). The backslash A\b finds an x with at most m nonzeros.
- Suppose A is tall and thin (m > n). The n columns are likely to be independent.
But if b is not in the column space, Ax = b has no solution. The least squares
method minimizes ‖b − Ax‖
2 by solving A
T Ax̂ = A
T b.
The Four Fundamental Subspaces
The nullspace N (A) contains all solutions to Ax = 0. The column space C (A)
contains all combinations of the columns. When A is m by n, N (A) is a subspace of
R
n and C (A) is a subspace of R
m .
The other two fundamental spaces come from the transpose matrix A
T
. They are
N (A
T ) and C (A
T ). We call C (A
T ) the “row space of A” because the rows of A are
the columns of A
T
. What are those spaces for our 2 by 2 example?
A = transposes to A
T
=.
3 6 2 6
Both columns of A
T are in the direction of (1, 2). The line of all vectors (c, 2 c) is
C (A
T ) = row space of A. The nullspace of A
T is in the direction of (3, −1):
Nullspace of A
T A
T y =
1 3 E
gives
E
3 c
.
2 6 F 0 F −c
The four subspaces N (A), C (A), N (A
T ), C (A
T ) combine beautifully into the big
picture of linear algebra. Figure A2 shows how the nullspace N (A) is perpendicular
to the row space C (A
T ). Every input vector x splits into a row space part x r
and a
nullspace part x n
. Multiplying by A always(!) produces a vector in the column space.
Multiplication goes from left to right in the picture, from x to Ax = b.
xr
x = xr + xn
x n
Axr
= b
Ax
b
Axn^
row space C (A
T )
multiples of (1, 2)
nullspace N (A)
multiples of (2, −1)
A =
[
]
column space
C (A)
multiples of (1, 3)
nullspace N (A
T )
multiples of (3, −1)
Figure A2: The four fundamental subspaces (lines) for the singular matrix A.
On the right side are the column space C (A) and the fourth space N (A
T ). Again
they are perpendicular. The columns are multiples of (1, 3) and the y’s are multi
ples of (3, −1). If A were an m by n matrix, its columns would be in m-dimensional
space R
m and so would the solutions to A
T y = 0. Our singular 2 by 2 example has
m = n = 2, and all four fundamental subspaces in Figure A2 are lines in R
2 .
This figure needs more words. Each subspace contains infinitely many vectors,
or only the zero vector x = 0. If u is in a space, so are 10 u and − 100 u (and most
importantly 0 u). We measure the dimension of a space not by the number of
vectors, which is infinite, but by the number of independent vectors. In this example
each dimension is 1. A line has one independent vector but not two
Dimension and Basis
A full set of independent vectors is a “basis” for a space. This idea is important.
The basis has as many independent vectors as possible, and their combinations fill
the space. A basis has not too many vectors, and not too few:
- The basis vectors are linearly independent.
- Every vector in the space is a unique combination of those basis vectors.
Here are particular bases for R
n among all the choices we could make:
Standard basis = columns of the identity matrix
General basis = columns of any invertible matrix
Orthonormal basis = columns of any orthogonal matrix
The “dimension” of the space is the number of vectors in a basis.
Difference Matrices
Difference matrices with boundary conditions give exceptionally good examples of the
four subspaces (and there is a physical meaning behind them). We choose forward
and backward differences that produce 2 by 3 and 3 by 2 matrices:
Forward Δ
Backward −Δ −
A =
[
]
and A
T
A is imposing no boundary conditions (no rows are chopped off). Then A
T must
impose two boundary conditions and it does: +1 disappeared in the first row and − 1
in the third row. A
T w = f builds in the boundary conditions w 0 = 0 and w 3 = 0.
The nullspace of A contains x = (1, 1 , 1). Every constant vector x = (c, c, c)
solves Ax = 0, and the nullspace N (A) is a line in three-dimensional space. The
row space of A is the plane through the rows (− 1 , 1 , 0) and (0, − 1 , 1). Both vectors
are perpendicular to (1, 1 , 1) so the whole row space is perpendicular to the
nullspace. Those two spaces are on the left side (the 3D side) of Figure A3.
