# Lesson: Linear Mappings Over \( \mathbb{C} \)

1 point

## Transcript

If V and W are vector spaces over the complex numbers, then a mapping L from V to W is a linear mapping if for any alpha in C, and v1 and v2 in V, we have that L((alpha)v1 + v2) = (alpha)(L(v1)) + L(v2).

Let’s look at some examples. The mapping L from C4 to C2 defined by L([z1; z2; z3; z4]) = [2z1 + 3z2; (2i)z3 + (3i)z4] is a linear mapping. We prove this as follows. We’ll note that L((alpha)z + w) = L([(alpha)z1 + w1; (alpha)z2 + w2; (alpha)z3 + w3; (alpha)z4 + w4]), which, using our definition of L, equals [2((alpha)z1 + w1) + 3((alpha)z2 + w2); (2i)((alpha)z3 + w3) + (3i)((alpha)z4 + w4)]. So now is when we want to expand out all of our terms. First, we’ll get [2(alpha)z1 + 2w1 + 3(alpha)z2 + 3w2; then (2i)(alpha)z3 + (2i)w3 + (3i)(alpha)z4 + (3i)w4]. Now we can sort of split things up into our z’s and our w’s, getting [2(alpha)z1 + 3(alpha)z2; (2i)(alpha)z3 + (3i)(alpha)z4] + [2w1 + 3w2; (2i)w3 + (3i)w4]. Now if we pull out our alpha from the first vector, we’ll see that this equals (alpha)[2z1 + 3z2; (2i)z3 + (3i)z4] + [2w1 + 3w2; (2i)w3 + (3i)w4], and then we’ll see that this equals (alpha)(L(z)) + L(w).

Let’s look at another example. The mapping M from C1 to C1 defined by M(z) = the conjugate of z is not a linear mapping, as it does not preserve scalar multiplication. For example, if we let alpha = 1 – 3i, v1 = 2 + 4i, and v2 = 0, then we see that M((alpha)v1 + v2) = M((1 – 3i)(2 + 4i) + 0). Well, this equals M(2 + 4i – 6i – 12(i-squared)), which equals M(14 – 2i), which will equal 14 + 2i. But now we calculate that (alpha)(M(v1)) + M(v2) would equal (1 – 3i)(M(2 + 4i)) + M(0), which equals (1 – 3i)(2 – 4i) + 0, which equals 2 – 4i – 6i + 12(i-squared), which is -10 – 10i. And so we have that M((alpha)v1 + v2) does not equal (alpha)(M(v1)) + M(v2).

As in Rn, we find that every linear mapping in Cn can be thought of as a matrix mapping. In our first example, we saw that L([z1; z2; z3; z4]) was equal to [2z1 + 3z2; (2i)z3 + (3i)z4]. Well, we can think of this as being the matrix [2, 3, 0, 0; 0, 0, 2i, 3i][z1; z2; z3; z4].

In general, we know that if z is a vector in Cn, then we can write that z = z1e1 + z2e2 + through to znen, where ei are the same as in Rn. Remember, every real number is a complex number, so every real vector is a complex vector. And so this means that we could write L(z) = L(z1e1 + z2e2 + through to znen), and then we could pull out our zj terms to get z1(L(e1)) + z2(L(e2)) + through to zn(L(en)), which we can write as the product of the matrix whose columns are [L(e1), L(e2), dot dot dot, L(en)] times our vector [z1; z2; through zn]. So, just as before, if we want to find the standard matrix [L] for a linear mapping L, we see that our matrix bracket-[L] equals the matrix whose columns are [L(e1) through L(en)].

Let’s look at an example. We’ll find the standard matrix for the linear mapping L from C3 to C3 defined by L([z1; z2; z3]) = [z1 + (1 – 3i)z3; (4i)z2; 2z1 + (1 – 5i)z2]. So first, we need to compute the following. Well, L(e1), which will equal L([1; 0; 0]), equals [1 + (1 – 3i)(0); (4i)(0); 2(1) + (1 – 5i)(0)], so this is the vector [1; 0; 2]. Now, L(e2) will equal L([0; 1; 0]). Well, this would equal [0 + (1 – 3i)(0); (4i)(1); then 2(0) + (1 – 5i)(1)], which is the vector [0; 4i; 1 – 5i]. Lastly, L(e3) will equal L([0; 0; 1]), which is [0 + (1 – 3i)(1); (4i)(0); 2(0) + (1 – 5i)(0)], so this equals [1 – 3i; 0; 0]. Remembering to turn these vectors into the columns of our standard matrix, we find that bracket-[L] equals the matrix [1, 0, 1 – 3i; 0, 4i, 0; 2, 1 – 5i, 0].