Lesson: Linear Mappings

Question 2

1 point

Transcript

Let’s return now to a concept we developed in our study of Rn: linear mappings. Because, as with bases, linear mappings will be of much greater use in the general world of vector spaces than they were in simply the Rn spaces. We will, of course, start with a definition.

If V and W are vector spaces over R, a function L from V to W is a linear mapping if it satisfies the linearity properties: L1, that L(x + y) = L(x) + L(y), and L2, that L(tx) = t(L(x)) for all x and y in V and t in the real numbers. If W = V, then L may be called a linear operator. Note that the two linearity properties can be combined into one statement, that L(tx + y) = t(L(x)) + L(y).

For example, the mapping L from 2-by-2 matrices to polynomials of degree less than or equal to 2 defined by L([a, b; c, d]) = (a + b) + (a + c)x + (a + d)(x-squared) is a linear mapping because L(tx + y), which would be equal to L(t[a1, b1; c1, d1] + [a2, b2; c2, d2]) = L(the matrix [ta1 + a2, tb1 + b2; tc1 + c2, td1 + d2]). Well, this will equal (ta1 + a2 + tb1 + b2) + (ta1 + a2 + tc1 + c2)x + (ta1 + a2 + td1 + d2)(x-squared). We can split this up and find that it equals (ta1 + tb1) + (ta1 + tc1)x + (ta1 + td1)(x-squared) + (a2 + b2) + (a2 + c2)x + (a2 + d2)(x-squared). Further pulling out our t, we get that this equals t times the quantity ((a1 + b1) + (a1 + c1)x + (a1 + d1)(x-squared)) and we still add our ((a2 + b2) + (a2 + c2)x + (a2 + d2)(x-squared)), which is to say that this equals t(L([a1, b1; c1, d1])) + L([a2, b2; c2, d2]), which is t(L(x)) + L(y).

On the other hand, the mapping M from P3 to P3 defined by M(a + bx + c(x-squared) + d(x-cubed)) = a-squared + (ab)x + (ac)(x-squared) + (ad)(x-cubed) is not a linear mapping. Consider, for example, that M(1 + 2x + x-squared + 2(x-cubed)) = 1 + 2x + x-squared + 2(x-cubed) and M(4 – x + 2(x-squared) – x-cubed) = 16 – 4x + 8(x-squared) – 4(x-cubed). So we have that M(1 + 2x + x-squared + 2(x-cubed)) + M(4 – x + 2(x-squared) – x-cubed) will equal (1 + 2x + x-squared + 2(x-cubed)) + (16 – 4x + 8(x-squared) – 4(x-cubed)), which equals 17 – 2x + 19(x-squared) – 2(x-cubed). But M((1 + 2x + x-squared + 2(x-cubed)) + (4 – x + 2(x-squared) – x-cubed)) will equal M(5 + x + 3(x-squared) + x-cubed), which is 25 + 5x + 15(x-squared) + 5(x-cubed). So we see that M(1 + 2x + x-squared + 2(x-cubed)) + M(4 – x + 2(x-squared) – x-cubed) does not equal M((1 + 2x + x-squared + 2(x-cubed)) + (4 – x + 2(x-squared) – x-cubed)). This means that M is not closed under addition, so it is not a linear mapping.

It happens that M is also not closed under scalar multiplication. Consider, for example, that M(1 + 2x + x-squared + 2(x-cubed)) = 1 + 2x + x-squared + 2(x-cubed), so 5(M(1 + 2x + x-squared + 2(x-cubed))) will equal 5(1 + 2x + x-squared + 2(x-cubed)), which is 5 + 10x + 5(x-squared) + 10(x-cubed). But M(5(1 + 2x + x-squared + 2(x-cubed))) will equal M(5 + 10x + 5(x-squared) + 10(x-to-the-cubed)), which is 25 + 50x + 25(x-squared) + 50(x-cubed). So we see that 5(M(1 + 2x + x-squared + 2(x-cubed))) does not equal M(5(1 + 2x + x-squared + 2(x-cubed))).

Now, before I set you off to work on problems on your own, you might be wondering how to tell at a glance if something is linear or not. My general advice is that if you only end up doing linear combinations with your entries—add, multiply by constant, for example—then it is probably linear. But if the definition involves multiplication, roots, or exponents, then the mapping is probably not linear. Again, this just helps you get a guess; you still need to prove it.

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