Matrices record where basis vectors go.
In our first brush with linear algebra, we only dealt with linear maps between the spaces spaces Rn for varying n. For those maps and those spaces, the convenient standard basis allowed us to record linear maps using the finite data of a matrix.
In this section we will see that a similar story plays out for maps between finite dimensional vector spaces: they too can be described by a matrix, but only after making a choice of “basis” on the domain and codomain.
Question 1 Let V be a vector space with basis (~v1, ~v2, ~v3) and W be a vector space with basis ( ~w1, ~w2).
Suppose there is a linear map L : V → W for which
L(~v1) = 3 ~w1+ 2 ~w2, L(~v2) = 3 ~w1− 2 ~w2, and L(~v3) = ~w1+ ~w2.
In light of all this, compute L(2~v1). But how will we write down our answer?
Where does L(2~v1) live?
Solution (a) In W . X (b) In V .
And since we know that L(2~v1) ∈ W , we’ll write our answer as α ~w1+ β ~w2 for some numbers α and β.
So say L(2~v1) = α ~w1+ β ~w2. Solution In this case, α = 6.
Solution And β = 4.
Next compute L(2~v1+ 3~v2− 4~v3) = α ~w1+ β ~w2.
53 Matrix of a linear map
Definition 3 Let L : V → W be a linear map between finite dimensional vector spaces, let BV = (~v1, ~v2, . . . , ~vn) be a basis for V , and let BW = ( ~w1, ~w2, . . . , ~wm) be a basis for W . Then L(~vi) = ai,1w~1+ ai,2w~2+ · · · + ai,mw~m.
Then the matrix with respect to the bases BV and BW is the matrix M whose entry in the ith column and jthrow is ai,j bases for R2 and R3, respectively.
Solution
Hint: The first column of the matrix will be 1 1
E2
but written with respect to the basis B2.
Remember the order of vectors in the basis matters.
Hint:
Hint: So the first column of the matrix is
Hint: Similarly, the second column is
Hint: So the matrix of this linear map is
What is the matrix for the linear map L x y
with respect to the bases B1 and B2?
53 Matrix of a linear map
Question 5 Let P2 be the space of polynomials of degree at most 2. Let B0 = (1, x, x2) and B1 = (1, (x − 1), (x − 1)2). Consider the map L : P2 → R given by L(p) = p(1).
Solution Hint:
L(1) = 1 L(x) = 1 L(x2) = 12= 1 Hint: The matrix of L with respect to B0 is1 1 1 What is the matrix of this linear map with respect to the basis B0? Solution
Hint:
L(1) = 1 L(x − 1) = 1 − 1 = 0 L((x − 1)2) = (1 − 1)2= 0 Hint: The matrix of L with respect to B1 is1 0 0 What is the matrix of this linear map with respect to the basis B1?
Question 6 Let P3 be the space of polynomials of degree at most 3. Let B = (1, x, x2, x3). Consider the map L : P3→ P3given by L(p(x)) = d
dxp(x). This map is linear (why?).
Solution
Hint: L is linear because the derivative of a sum of two functions is the sum of the derivative of the two functions, and since the derivative of a constant times a function is the constant times the derivative of the function.
Hint:
54 Subspaces
A subspace is a subset of a vector space which is also a vector space.
Definition 1 A subset U of a vector space V is a subspace of V if U is a vector space with respect to the scalar multiplication and the vector addition inherited from V .
Question 2 Which of the following is a subspace of R2? Solution
Hint: The vectors1 2
and0
1
are both on the line `, but the sum1 2
+0
1
is not on the line `.
Hint: So ` is not a subspace.
Hint: The set P consists of a single vector.
Hint: But the vector in P is not the origin0 0
.
Hint: So1 2
∈ P but 10 ·1 2
6∈ P .
Hint: So P is not a subspace.
Hint: By process of elimination, the x-axis must be a subspace. Is it really?
Hint: Yes, if I multiply any vectorx 0
by a scalar, it is still on the x-axis.
Hint: And if I add together two vectors of the form x 0
, the result is still on the x-axis.
(a) The x-axis. X (b) The set P =1
2
.
(c) The line ` =x y
∈ R2 : y = x + 1
.
Let’s look at some more examples! Which of the following is a subspace of R2? Solution
54 Subspaces
Hint: The set A is not a subspace because1 0
∈ A, but −1 ·1 0
is not in A, so a scalar multiple of something in A need not be in A.
Hint: The set C is not a subspace because even though it is closed under scalar multiplication (check this!) it is not closed under vector addition, since 1
−2
and1
2
are both in C, but their sum2
0
is not (draw a picture of this example!).
Hint: As the only choice left, B must be a subspace.
The reason is that it is just the span of the vector2 1
, and as such, is closed under scalar multiplication and vector addition.
(a) The set A = {x y
: x > 0 and y > 0}
(b) The set B =x y
: x = 2y X
(c) The set C = {x y
: |y| < |x|}
Solution
Hint: Question 3 What about the line y = 3? Does it form a subspace of R2? Solution
(a) Yes.
(b) No. X
That’s right; the tip of the vector0 3
is on that line, but the scalar multiple of that vector, like 2 ·0
3
=0 6
, is not on the line.
So when do the points on a line in R2 form a subspace?
55 Kernel
A kernel is everything sent to zero.
There are some special subspaces that we will want to pay attention to.
Theorem 1 If L : V → W is a linear transformation, then the kernel of L, defined by
ker(L) = {~v ∈ V : L(~v) = ~0}
is a subspace of V .
You may also hear this referred to as the null space of L.
Prove this theorem that ker L is a subspace.
We only need to show that ker(L) is closed under scalar multiplication and vector addition.
For any ~v ∈ ker(L) and c ∈ R,
L(c~v) = cL(~v)
= c~0
= ~0 so c~v ∈ ker(L).
If ~v, ~w ∈ ker(L), then
L(~v + ~w) = L(~v) + L( ~w)
= ~0 + ~0
= ~0 so ~v + ~w ∈ ker(L)
Thus ker(L) is a subspace!
Question 2 Let L : R3→ R2be the linear map whose matrix is2 3 1 1 0 −1
Solution
Hint: Just by evaluating all three, we see the only one which gets sent to0 0
by L is
1
−1 1
Which of the following vectors is in the kernel of L?
(a)
1
−1 1
X
(b)
3 2 0
55 Kernel
(c)
0 0 2
Theorem 3 A linear map L : V → W is injective if and only if ker(L) = {~0}.
Definition 4 The word “injective” is an adjective meaning the same thing as
“one to one.” In other words, a function f : A → B is injective if f (a1) = f (a2) implies a1= a2.
Prove this theorem.
Let L be injective. Then L(~v) = ~0 implies L(~v) = L(~0). Since L is injective, this implies ~v = ~0. Thus the only element of the kernel is ~0.
On the other hand, if ker(L) = {~0}, then if L( ~v1) = L( ~v2), then L( ~v1− ~v2) = ~0, so ~v1− ~v2 is in the null space, and hence must be equal to ~0. But then we can conclude that ~v1= ~v2
Definition 5 The dimension of the kernel of L is the nullity of L.
Be careful to observe that ker L is a subspace, while dim ker L is a number, so the nullity of L is just a number.
56 Image
The “image” is every actual output.
Definition 1 If L : V → W is a linear transformation, then the image of L is Imag(L) = { ~w ∈ W : ∃~v ∈ V, L(~v) = ~w}
.
Remember to read ∃ as “there exists.”
Warning 2 Some people may call this the “range.” Some other people use the word “range” for what we’ve been calling the codomain. The result is that, in my opinion, the word “range” is now overused, so we give up and never use the word.
Question 3 Suppose L : R2→ R3, and suppose that
Hint: Use the fact that
Hint: In other words,
Hint: And so a vector in the domain which is sent to
This is a special case of a general fact: if we have two vectors in the image, then their sum is in the image, too.
Theorem 4 The image of a linear map is a subspace of the codomain.
56 Image
Prove this.
If w ∈ Imag(L), then there is a v ∈ V with L(v) = w. L(cv) = cL(v) = cw, so cw ∈ Imag(L) for any c ∈ R. Thus Imag(L) is closed under scalar multiplication.
If w1, w2∈ Imag(L), then there are v1, v2∈ V with L(v1) = w1and L(v2) = w2. L(v1+ v2) = L(v1) + L(v2) = w1+ w2, so w1+ w2 ∈ Imag(L). Thus Imag(L) is closed under vector addition.
We finish with some terminology.
Definition 5 The dimension of the image of L is the rank of L.
Be careful to observe that the image of L is a subspace, while the dimension of the image of L is a number, so the rank of L is just a number.
Question 6 Consider the linear map L : R2→ R3given by the matrix
2 1 4 2 6 3
. Solution
Hint: Not every vector in R3 is the image of L.
Hint: Let’s think about which vectors are in the image of L.
Question 7 Is
2 4 6
in the image of L?
Solution (a) Yes. X (b) No.
In fact,
2 4 6
= L1 0
.
Is
1 1
in the image of L?
56 Image
Solution (a) 0 (b) 1 X (c) 2
And so the rank is one.
The rank of L is 1.