Lesson: Complex Vector Spaces

Question 2

1 point

Transcript — Introduction

In this lecture, we will extend almost everything we did in MATH 136 with real vector spaces to the complex case.

Definition: A set V is called a vector space over C, or a complex vector space, if there is an operation of addition and an operation of scalar multiplication such that, for any vectors v, z, and w in V and scalars alpha and beta in C, we have the following ten properties.

Notice that a vector space over C has exactly the same definition as a vector space over R except that we now allow the scalar numbers to be complex numbers. All of our concepts and definitions of subspaces, spanning, linear independence, bases, dimensions, and coordinates are also exactly the same except that we now allow the use of complex scalars. As a result, all of our theory and algorithms are exactly the same.

I will demonstrate this with a few examples, but first, an important definition.

Definition: The set Cn of all vectors of the form [z1 to zn] such that zi is a complex number, with addition defined by the vector [z1 to zn] + [w1 to wn] = [(z1 + w1) up to (zn + wn)] and complex scalar multiplication defined by (alpha)[z1 to zn] = [(alpha)z1 to (alpha)zn] for any complex number alpha, is a vector space over C. This is, of course, the complex vector space version of Rn, and will be the main complex vector space that we use in this course.

As we did in the real case, the first thing we should think about when we have a new vector space is what is a basis and the dimension of the vector space. Notice that since we allow complex scalar multiplication, we have that a basis for Cn is {e1 to en}—i.e., the same standard basis as we had for Rn. And so, the dimension of Cn is equal to n. So when I say that Cn is the complex version of Rn, I really mean it. It has the same standard basis and the same dimension.

Note that, just like a complex number z in C, we can split a vector in Cn into its real and imaginary parts. Theorem 11.2.1: If z is a vector in Cn, then there exists vectors x and y in Rn such that z = x + iy.


Okay, now on to the examples. Example: Is the set B = {[1; i; 1 + i], [2i; -1; 1], [1; 1; i]} a basis for C3? Solution: We solve this the same way we did problems back in MATH 136 in R3. So, for linear independence, we consider the 0 vector = (alpha1)[1; i; 1 + i] + (alpha2)[2i; -1; 1] + (alpha3)[1; 1; i]. Calculating the linear combination on the right-hand side and comparing entries gives a homogeneous system of linear equations. Row reducing the corresponding coefficient matrix gives the reduced row echelon form is the identity matrix. Thus, the system has a unique solution alpha1 = alpha2 = alpha3 = 0, and so the set B is linearly independent. Thus, B is a linearly independent set of 3 vectors in a 3-dimensional vector space, and hence, B is a basis for C3.

Example: Find the coordinates of z = [2i; -2 + i; 3 + 2i] with respect to the basis B = {[1; 1 + i; 1 – i], [-1; -i; -1 + 2i], [i; i; 2 + 2i]} for C3. Solution: By definition of coordinates, we need to find complex numbers alpha1, alpha2, and alpha3 such that [2i; -2 + i; 3 + 2i] = (alpha1)[1; 1 + i; 1 – i] + (alpha2)[-1; -i; -1 + 2i] + (alpha3)[i; i; 2 + 2i]. Calculating the linear combination on the right-hand side and comparing entries gives a complex system of linear equations. Row reducing, we get the reduced row echelon form [1, 0, 0 | i; 0, 1, 0 | -i; 0, 0, 1 | 0]. And thus, the coordinate vector of z with respect to the basis B is [i; -i; 0].

Example: Is Rn a subspace of Cn? Solution: As we saw with real vector spaces, to prove a set is a subspace of another vector space, we use the Subspace Test. First, it is clear that every vector x = [x1 to xn] in Rn is in Cn since every real number xi is a complex number with 0 imaginary part. Thus, Rn is a non-empty subset of Cn. Next, if we take any two vectors x and y in Rn, then we know that x + y is in Rn since we already know that Rn is closed under standard addition. Finally, for any scalar t, we have that tx does not have to be in Rn. Again, this is the big difference between real and complex vector spaces—we now allow multiplication by complex scalars. So, for example, the vector e1 is in Rn, but (2 + 3i)e1 is not in Rn, so Rn is not closed under complex scalar multiplication, and so it is not a vector space over C. Therefore, Rn is not a subspace of Cn.

The other complex vector space that we will refer to in this course is the extension of M(m-by-n)(R). Definition: The set M(m-by-n)(C) of all m-by-n matrices with complex entries is a complex vector space with standard addition and complex scalar multiplication of matrices.

We can also extend all of our definitions and theorems about linear mappings, including isomorphisms, to complex vector spaces.

Example: Find the standard matrix of the linear mapping L from C2 to C3 given by L(z1, z2) = (iz1, z2, z1 + z2). Solution: First, what is the standard basis for C2? It is just the same standard basis as for R2. That is, it is {[1; 0], [0; 1]}. Thus, we have L of the first standard basis vector, by definition of the mapping, is [i; 0; 1]. L of the second standard basis vector is [0; 1; 1]. And so the standard matrix of L is the matrix whose columns are the images of the standard basis vectors under L, and so it is [i, 0; 0, 1; 1, 1].

Example: Is the mapping L from C to C defined by L(z1) = the complex conjugate of z1 a linear mapping? Solution: We try to prove this the same way we would have back in MATH 136. For any vectors z1 and z2 in C, and complex scalars alpha1 and alpha2 in C, consider L((alpha1)z1 + (alpha2)z2). By definition of the mapping, this is equal to the complex conjugate of ((alpha1)z1 + (alpha2)z2), which is equal to, by properties of the complex conjugate, (the conjugate of alpha1)(the conjugate of z1) + (the conjugate of alpha2)(the conjugate of z2). But notice, the complex conjugate of a complex number is not equal to itself if the complex number is not real, and so we suspect that this mapping is not linear. To prove it is not linear, we need to provide a counterexample. Taking alpha = (1 + i) and z = 1, we get L((alpha)z) = L((1 + i)(1), which is equal to the conjugate of ((1 + i)(1)), which is 1 – i. But (alpha)L(z) = (1 + i)L(1) = 1 + i. And so since L((alpha)z) does not equal (alpha)L(z), L is not linear.

We also directly extend the concepts of determinants and inverses.

Example: Find the inverse of Z = [-2i, 1 – i; 1 – 3i, 2]. Solution: We have the determinant of Z is equal to (-2i)(2) – (1 – i)(1 – 3i). Calculating this, we get it’s equal to 2. Thus, from our formula for the inverse of a 2-by-2 matrix, we get Z-inverse = (1/2)[2, -1 + i; -1 + 3i, -2i].

We now define one new operation on Cn and M(m-by-n)(C). Definition: For any vector z = [z1 to zn] in Cn and any matrix Z = [z11 to z1n; …; zm1 to zmn] in M(m-by-n)(C), we define the complex conjugate by the complex conjugate of the vector z is obtained by taking the complex conjugate of every entry of the vector z, and similarly, the complex conjugate of the matrix Z we get by taking the complex conjugate of each entry of the matrix.

Example: Let z be the vector [1 – 3i; -3; 2; 4i] and Z be the matrix [3, -2i; 1 + 3i, 1 – i]. Then the conjugate of the vector z is [1 + 3i; -3; 2; -4i] and the conjugate of the matrix Z is [3, 2i; 1 – 3i, 1 + i].

Finally, I will conclude this lecture by stating one theorem about the complex conjugate of matrices and vectors that we will require a few times. Theorem 11.2.2: If A is in M(m-by-n)(C) and z is in Cn, then the conjugate of (Az) is equal to (the conjugate of A)(the conjugate of z).

This ends this lecture.

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