Summarizing these results: Row operations change neither the row space nor the null space of A.
Corollary 1: If R is the Gauss-Jordan form of A, then R has the same null space and row space asA.
Corollary 2: IfB ∼A, thenR(B) =R(A), and N(B) =N(A).
Proof: If B ∼ A, then both A and B have the same GJ form, and hence the same rank (equal to the number of leading ones) and nullity (equal to the number of free variables).
The following result may be somewhat surprising:
Theorem: The number of linearly independent rows of the matrix A is equal to the number of linearly independent columns of A. In particular, the rank of A is also equal to the number of linearly independent columns, and hence to the dimension of the column space of A
Proof (sketch): As an example, consider the matrix
A=
3 1 −1
4 2 0
2 3 4
Observe that columns 1, 2, and 3 are linearly dependent, with col1(A) = 2col2(A)−col3(A).
You should be able to convince yourself that doing any row operation on the matrix A doesn’t affect this equation. Even though the row operation changes the entries of the various columns, it changes them all in the same way, and this equation continues to hold.
The span of the columns can, and generally will change under row operations (why?), but this doesn’t affect the result. For this example, the column space of the original matrix has dimension 2 and this is preserved under any row operation.
The actual proof would consist of the following steps: (1) identify a maximal linearly in-dependent set of columns of A, (2) argue that this set remains linearly independent if row operations are done on A. (3) Then it follows that the number of linearly independent columns in the GJ form of A is the same as the number of linearly independent columns in A. The number of linearly independent columns of A is then just the number of leading entries in the GJ form of A which is, as we know, the same as the rank of A.
where n is the number of columns of A.
Let’s assume, for the moment, that this is true. What good is it? Answer: You can read off both the rank and the nullity from the echelon form of the matrix A. Suppose A can be
row-reduced to
1 ∗ ∗ ∗ ∗ 0 0 1 ∗ ∗ 0 0 0 1 ∗
.
Then it’s clear (why?) that the dimension of the row space is 3, or equivalently, that the dimension of the column space is 3. Since there are 5 columns altogether, the dimension theorem says that n = 5 = 3 +N(A), so N(A) = 2. We can therefore expect to find two linearly independent solutions to the homogeneous equation Ax=0.
Alternatively, inspection of the echelon form of A reveals that there are precisely 2 free variables,x2 and x5. So we know thatN(A) = 2 (why?), and therefore, rank(A) = 5−2 = 3.
Proof of the theorem: This is, at this point, almost trivial. We have shown above that the rank of A is the same as the rank of the Gauss-Jordan form of A which is equal to the number of leading entries in the Gauss-Jordan form. We also know that the dimension of the null space is equal to the number of free variables in the reduced echelon (GJ) form ofA.
And we know further that the number of free variables plus the number of leading entries is exactly the number of columns. So
n =N(A) +R(A), as claimed.
♣ Exercise:
• Find the rank and nullity of the following - do the absolute minimum (zero!) amount of computation possible:
3 1
−6 −2
,
2 5 −3
1 4 2
• (T/F) For any matrix A, R(A) = R(At). Give a proof or counterexample.
• (T/F) For any matrix A, N(A) =N(At). Give a proof or counterexample.
Chapter 14 Change of basis
When we first set up a problem in mathematics, we normally use the most familiar coordi-nates. InR3, this means using the Cartesian coordinates x, y, and z. In vector terms, this is equivalent to using what we’ve called the standard basis inR3; that is, we write
x y z
=x
1 0 0
+y
0 1 0
+z
0 0 1
=xe1+ye2+ze3,
where {e1,e2,e3} is the standard basis.
But, as you know, for any particular problem, there is often another coordinate system that simplifies the problem. For example, to study the motion of a planet around the sun, we put the sun at the origin, and use polar or spherical coordinates. This happens in linear algebra as well.
Example: Let’s look at a simple system of two first order linear differential equations dx1
dt = 3x1+x2 (14.1)
dx2
dt = x1+ 3x2.
To solve this, we need to find two functions x1(t), and x2(t) such that both equations hold simultaneously. Now there’s no problem solving a single differential equation like
dx/dt= 3x.
In fact, we can see by inspection thatx(t) =ce3t is a solution for any scalarc. The difficulty with the system (1) is that x1 and x2 are ”coupled”, and the two equations must be solved simulataneously. There are a number of straightforward ways to solve this system which you’ll learn when you take a course in differential equations, and we won’t worry about that
But there’s also a sneaky way to solve (1) by changing coordinates. We’ll do this at the end of the lecture. First, we need to see what happens in general when we change the basis.
For simplicity, we’re just going to work inR2; generalization to higher dimensions is (really!) straightforward.
14.1 The coordinates of a vector
Suppose we have a basis {e1,e2} forR2. It doesn’t have to be the standard basis. Then, by the definition of basis, any vectorv∈R2can be written as a linear combination ofe1 and e2. That is, there exist scalars c1, c2 such that v=c1e1+c2e2.
Definition:
The numbers c1 and c2 are called the coordinates of v in the basis {e1,e2}. Andve = c1
c2
is called the coordinate vectorof vin the basis {e1,e2}. Theorem: The coordinates of the vector v areunique.
Proof: Suppose there are two sets of coordinates forv. That is, suppose thatv=c1e1+c2e2, and also thatv=d1e1+d2e2. Subtracting the two expressions for v gives
0= (c1−d1)e1 + (c2−d2)e2.
But {e1,e2} is linearly independent, so the coefficients in this expression must vanish: c1− d1 =c2 −d2 = 0. That is,c1 =d1 and c2 =d2, and the coordinates are unique, as claimed.
Example: Let us use the basis
{e1,e2}= 1
2
,
−2 3
,
and suppose
v= 3
5
.
Then we can find the coordinate vectorvein this basis in the usual way, by solving a system of linear equations. We are looking for numbers c1 and c2 (the coordinates of vin this basis) such that
c1
1 2
+c2
−2 3
= 3
5
.
In matrix form, this reads
Ave =v, where
A=
1 −2
2 3
, v=
3 5
, and ve= c1
c2
.
We solve for ve by multiplying both sides by A−1: ve =A−1v= (1/7)
3 2
−2 1
3 5
= (1/7) 19
−1
=
19/7
−1/7
♣ Exercise: Find the coordinates of the vector v= (−2,4)t in this basis.