## Transcript

There are two subspaces that we affiliate with linear mappings: the range and the nullspace. We will define them first, and then prove that they are subspaces.

Let V and W be vector spaces over R. The range of a linear mapping L from V to W is defined to be the set of all L(x) in W for x in V.

For example, let’s let L be the mapping from R2 to M(2, 2) defined by L([a; b]) = [a, b; b, a]. Then we see that [1, 2; 2, 1] is in the range of L because L([1; 2]) = [1, 2; 2, 1]. But [1, 2; 3, 4] is not in the range of L. To see this, we note that [1, 2; 3, 4] is in the range of L only if there is a vector [a; b] whose entries a and b satisfy that L([a; b]), which equals [a, b; b, a], equals [1, 2; 3, 4]. Setting the matrix entries equal to each other, we see that this is equivalent to the system a = 1, b = 2, b = 3, a = 4. But since we cannot simultaneously have a = 1 and a = 4 (or b = 2 and b = 3), we see that our system is inconsistent. And since there are no such a and b, we have that [1, 2; 3, 4] is not in the range of L.

For another example, let’s determine whether or not the vector y = [5; -7] is in the range of the mapping L from P3 to R2, where L is defined by L(a + bx + c(x-squared) + d(x-cubed)) = [a + 2b + c + 2d; 3a + 4b – c – 2d]. If y is in the range, find a vector x such that L(x) = y. First, we look to see if [5; -7] is in the range of L. That is, we want to know if there is a polynomial x = a + bx + c(x-squared) + d(x-cubed) whose coefficients a, b, c, and d satisfy that L(x), which equals [a + 2b + c + 2d; 3a + 4b – c – 2d] equals our vector y, [5; -7]. Setting the components equal to each other, we see that this is equivalent to this system. We solve the system by row reducing its augmented matrix, which only takes one step. Our matrix is now in row echelon form, and since there are no bad rows, we know that our system is consistent. This means that there is a solution, and thus, that [5; -7] is in the range of L.

So now let’s move on to finding an x such that L(x) = y. Well, this simply means finding a solution to our system, so we will continue our row reduction until we reach reduced row echelon form. This only takes two more steps. We see that our system is equivalent to the system a – 3c – 6d = -17, and b + 2c + 4d = 11. Replacing the variable c with the parameter s and the variable d with the parameter t, we see that the general solution to our system is [-17; 11; 0; 0] + s[3; -2; 1; 0] + t[6; -4; 0; 1]. Well, this provides us a list of all possible a, b, c, and d. By picking specific values for s and t, we can get a vector x such that L(x) = y, and the easiest possible values for s and t are s = 0, t = 0. This gives us that a = -17, b = 11, c = 0, and d = 0. And so we’ve seen that L(-17 + 11x) equals the vector [5; -7].

We now define the nullspace. The nullspace of L is the set of all vectors in V whose image under L is the 0 vector. We write in set notation that the Null(L) is the set of {x in V such that L(x) = 0}.

For example, let L be the linear mapping from M(2, 3) to P1 defined by L([a, b, c; d, e, f]) = (a + b + c) + (d + e + f)x. Well, then the matrix [1, 1, -2; -3, 0, 3] is in the nullspace of L since L([1, 1, -2; -3, 0, 3]) will equal (1 + 1 – 2) + (-3 + 0 + 3)x, which equals 0 + 0x, our 0 polynomial. But the matrix [1, 2, 1; -3, 5, 3] is not in the nullspace of L since L([1, 2, 1; -3, 5, 3]) will equal (1 + 2 + 1) + (-3 + 5 + 3)x, which equals 4 + 5x, which is not equal to our 0 polynomial.

You’ll note that it is much easier to check if a vector is in the nullspace than to check if it is in the range.

Now recall that one of the reasons we undertook a study of general vector spaces instead of focusing on each one individually is that it gives us the opportunity to prove statements that are true for all vector spaces—or, as below, a statement that is true for all linear mappings.

Theorem 4.5.1: Let V and W be vector spaces, and let L from V to W be a linear mapping. Then the following are true.

- L of our V 0 vector equals our W 0 vector.
- The nullspace of L is a subspace of V.
- The range of L is a subspace of W.

To prove part 1, let x be any element of V. Then we have that L of our V 0 vector equals L(0x) because, of course, we know that 0x always equals the 0 vector. By the linearity properties of L, we can pull the 0 out and see that this is equal to 0(L(x)). But again, 0 times any vector is 0, so this must equal our 0 vector in W.

I will leave the proof of part 2 as an assignment, and instead, I will now show you the proof of part 3. To see that the range of L is a subspace of W, we first want to note that the range of L is explicitly defined as a subset of W. Moreover, the range of L is non-empty since part 1 of our theorem tells us that L of the 0 vector will equal the 0 vector, and thus, we know for fact that our 0 vector is in the range. Next, we check to see if Range(L) is closed under addition. So let’s let w1 and w2 be elements of the range of L. Well, this means that there are elements v1 and v2 of V such that L(v1) = w1 and L(v2) = w2. Then we note that L(v1 + v2) will equal L(v1) + L(v2), which equals w1 + w2. So, v1 + v2 is an element of V such that L(v1 + v2) = w1 + w2, and thus w1 + w2 is in the range of L.

Well, now that we have shown that Range(L) is closed under addition, let’s show it is closed under scalar multiplication. To that end, we’ll let w be in the range of L and s be a scalar. Well, then there is a v in V such that L(v) = w. But this means that L(sv) = s(L(v)), which equals sw. So we have found that sv is an element of V such that L(sv) = sw, so sw is in the range of L.