Lesson: The Rank-Nullity Theorem

Question 2

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Transcript

Since the Range(L) and the Null(L) are subspaces of W and V respectively, we can try to find a basis for them.

For our first example, let’s determine a basis for the range and nullspace of the linear mapping L from R2 to M(2, 2) defined by L([a; b]) = [a, b; b, a]. Well, the range of L is matrices of the form [a, b; b, a], which we could write as a[1, 0; 0, 1] + b[0, 1; 1, 0]. So we see that the set B = {[1, 0; 0, 1], [0, 1; 1, 0]} is a spanning set for the range of L. But moreover, the matrices of B are clearly not a scalar multiple of each other, so B is linearly independent, too. So B must be a basis for the range of L.

Now, the nullspace of L consists of all vectors [a; b] whose entries a and b satisfy that L([a; b]), which equals [a, b; b, a], equals [0, 0; 0, 0]. Setting the entries equal to each other, we see that this means that a = 0 and b = 0. So the only element of the nullspace is the vector [0; 0]. That is, Null(L) is simply {the 0 vector}, and the basis for this is the empty set.

Now let’s determine a basis for the range and nullspace of the linear mapping L from P3 to R2 defined by L(a + bx + c(x-squared) + d(x-cubed)) = [a + 2b + c + 2d; 3a + 4b – c – 2d]. Well, we can write the range of L as vectors of the form [a + 2b + c + 2d; 3a + 4b – c – 2d], which is a[1; 3] + b[2; 4] + c[1; -1] + d[2; -2]. And so we see that the set B = {[1; 3], [2; 4], [1; -1], [2; -2]} is a spanning set for the range of L. But is it linearly independent? Well, to check, we need to look for solutions to the equation [0; 0] = a[1; 3] + b[2; 4] + c[1; -1] + d[2; -2]. That is to say, the 0 vector equals [a + 2b + c + 2d; 3a + 4b – c – 2d]. Setting the entries equal to each other, we see that this is equivalent to this system, and we solve this homogeneous system by row reducing its coefficient matrix. From the row echelon form, we see that the solution to our system will have two parameters, so B is linearly dependent. However, since B is a spanning set, we know that we can remove the dependent members in order to end up with a basis. To that end, we will continue our row reduction to find the solution to the homogeneous system, and use the solution to write elements of B as a linear combination of the other elements. So we continue our row reduction, and we see that our system is equivalent to the system a – 3c – 6d = 0, b + 2c + 4d = 0. 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 this: s[3; -2; 1; 0] + t[6; -4; 0; 1]. Now if we set s = -1 and t = 0, this means that (-3)[1; 3] + 2[2; 4] – [1; -1] + 0[2; -2] = [0; 0], and so we see that [1; -1] = (-3)[1; 3] + 2[2; 4]. Meanwhile, setting s = 0 and t = -1 gives us that (-6)[1; 3] + 4[2; 4] + 0[1; -1] – 1[2; -2] = [0; 0], so we have that [2; -2] = (-6)[1; 3] + 4[2; 4]. And so we can remove both [1; -1] and [2; -2] from B to get the set B1 = {[1; 3], [2; 4]}, which is still a spanning set for the range of L. And since the vectors in B1 are not scalar multiples of each other, B1 is also linearly independent, and so we see that B1 is a basis for the range of L.

The nullspace of L consists of all polynomials a + bx + c(x-squared) + d(x-cubed) whose coefficients satisfy that L(a + bx + c(x-squared) + d(x-cubed)), which equals [a + 2b + c + 2d; 3a + 4b – c – 2d], equals the 0 vector, [0; 0]. Well, luckily, we’ve already solved this equation in our efforts to find the range of L. The solution was s[3; -2; 1; 0] + t[6; -4; 0; 1]. Plugging this solution in, we see that the nullspace of L consists of all polynomials of the form s(3 – 2x + x-squared) + t(6 – 4x + x-cubed). And so we see that the set C = {3 – 2x + x-squared, 6 – 4x + x-cubed} is a spanning set for the nullspace of L. And since the polynomials are not scalar multiples of each other, C is clearly linearly independent as well, and so we have that C is a basis for the nullspace of L.

Now, once we have a basis for the range or the nullspace for a linear mapping, we can calculate its dimension. It turns out the dimension of these special subspaces is something we want to concern ourselves with, so we give them their own special name.

Let V and W be vector spaces over R. The rank of a linear mapping L from V to W is the dimension of the range of L.

Let V and W be vector spaces over R. The nullity of a linear mapping L from V to W is the dimension of the nullspace of L.

So continuing our examples, in our first example, we found that the set {[1, 0; 0, 1], [0, 1; 1, 0]} is a basis for the range of L. Since this basis has two matrices in it, we see that the rank of L is 2. We then found that the empty set is a basis for the nullspace of L, and since the empty set has zero entries in it, the nullity of L is 0.

In our second example, we found that the set {[1; 3], [2; 4]} is a basis for the range of L. Since this set has two elements, the rank of L is 2. We then found that the set {3 – 2x + x-squared, 6 – 4x + x-cubed} is a basis for the nullspace of L. Since this set has two elements, the nullity of L is also 2.

Now, note that in our first example, we got that the rank of L + the nullity of L = 2 + 0, which equals 2, which is the dimension of R-squared. We also see that, in our second example, the rank of L + the nullity of L = 2 + 2, which equals 4, which is the dimension of P3. That is, in both of these examples, the sum of the rank and the nullity equals the dimension of the domain. As you may be guessing, this is not a coincidence.

Theorem 4.5.2, the Rank-Nullity Theorem: Let V and W be vector spaces over R, with the dimension of V equal to n, and let L from V to W be a linear mapping. Then the rank of L + the nullity of L = n.

To prove the Rank-Nullity Theorem, we are, in fact, going to show that the rank of L equals the dimension of V minus the nullity of L by finding a basis for the range of L with n – nullity(L) elements in it. To that end, let’s say that the nullity of L is k, and that the set {v1 through vk} is a basis for the nullspace of L. Now, since the nullspace of L is a subspace of V, we can expand the linearly independent set {v1 through vk} into a basis for V.

Let’s let {v1 through vk, u(k+1) through un} be such a basis. So first, let’s consider the possibility that the {v1 through vk} may have already been a spanning set for V, so that k = n, and no additional ui vectors are necessary. Well, in this case, it would mean that every vector is mapped by L to the 0 vector, which means that the only vector in the range of L is the 0 vector. Thus, the range of L is the single-element set which has the empty set for a basis, and so we see that the rank of L is 0. And this means that the rank of L + the nullity of L = 0 + n, which equals n, as desired.

Otherwise, k is less than n, and at least one ui is necessary. Now let’s let w be any vector in the range of L. Well, then there’s some vector x in V such that L(x) = w. But since x is in V, we know it can be written as a linear combination of our basis elements. That is, we have scalars t1 through tn such that x = t1v1 + dot dot dot + tkvk + t(k+1)u(k+1) + dot dot dot + tnun. And since w = L(x), we have that w = L of this enormous sum, but we can use our linearity properties to see that this equals t1(L(v1)) + through to tk(L(vk)) + t(k+1)(L(u(k+1))) + through to tn(L(un)). But remember, our vectors v1 through vk are in the nullspace, so that means that L sends them to 0. So this is really equal to t1(0) + through to tk(0) + t(k+1)(L(u(k+1))) + through to tn(L(un)), but since all those initial terms go to 0, we see that w = t(k+1)(L(u(k+1))) through to tn(L(un)). So we see that every element w in the range of L can be written as a linear combination of the vectors L(u(k+1)) through L(un). {L(u(k+1)) through L(un)} is a spanning set for the range of L.

But is it linearly independent? Well, suppose that we had scalars s(k+1) through sn such that s(k+1)(L(u(k+1))) + all the way through to sn(L(un)) equaled the 0 vector. Well, using the linearity of L, we would get that L(s(k+1)u(k+1) + all the way through to snun) equals our 0 vector. Well, this means that s(k+1)u(k+1) + all the way through to snun is in the nullspace of L. This means that we could write s(k+1)u(k+1) + all the way through snun as a linear combination of the elements of our basis for the nullspace of L, {v1 through vk}. So, there are scalars s1 through sk such that our vector s(k+1)u(k+1) + all the way through to snun equals the linear combination s1v1 + all the way through to skvk. However, this means that we can write –s1v1 – through – skvk + s(k+1)u(k+1) + through to snun equal to the 0 vector. Well, since our set is a basis for V, it is linearly independent, so the only solution to this equation is for s1 through sn = 0. Well, this means specifically that s(k+1) through sn will have to equal to 0 is the only solution to s(k+1)(L(u(k+1))) + all the way through to sn(L(un)) equals our 0 vector. And so we’ve shown that our set {L(u(k+1)) through L(un)} is linearly independent, and thus is a basis for the range of V.

This means that the rank of L is n – k, and we have shown that the rank of L + the nullity of L = (n – k) + k, which equals n.

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