Complex Vector Spaces

Introduction

We have already looked at \(\mathbb{R}^{n}\) in considerable detail in earlier courses. We will now briefly look at \(\mathbb{C}^{n}\). Rather than working with a general value of \(n\), let us try to understand \(\mathbb{C}^{2}\). All the ideas can be easily extended to the case of \(n >2\).

\(\mathbb{R}^{2}\) is the set of all tuples of the form \(( x,y)\) where both \(x\) and \(y\) are real numbers. We denote this formally as:

\[ \begin{equation*} \mathbb{R}^{2} =\{( x,y) \ |\ x,y\in \mathbb{R}\} \end{equation*} \] We are quite familiar with \(\mathbb{R}^{2}\). Geometrically, it denotes the 2D plane. Extending this idea, the set \(\mathbb{C}^{2}\) is the set of all tuples of the form \(( z_{1} ,z_{2})\) where both \(z_{1}\) and \(z_{2}\) are complex numbers. Formally:

\[ \begin{equation*} \mathbb{C}^{2} =\{( z_{1} ,z_{2}) \ |\ z_{1} ,z_{2} \in \mathbb{C}\} \end{equation*} \] Some of the elements of \(\mathbb{C}^{2}\) are as follows:

• \(( i,2i)\)
• \(( 1+3i,-4+11i)\)
• \(( -2,3)\)
• \(( i,5)\)
• \(( 3,i)\)
• \(( 0,0)\)

Trying to gain a geometric understanding of \(\mathbb{C}^{2}\) is going to be hard. We would have to rely on our visual understanding of \(\mathbb{R}^{2}\) and try to extraploate it to \(\mathbb{C}^{2}\).

Complex Vector Space

Recall that \(\mathbb{R}^{2}\) can be viewed as a vector space if we add additional structure to it, namely vector addition and scalar multiplication. In a similar sense, \(\mathbb{C}^{2}\) can also be viewed as a vector space. One notable fact is that \(\mathbb{C}^{2}\) is a complex vector space. The scalars used to multiply vectors will now be drawn from \(\mathbb{C}\). Most of the vector operations that we learned in \(\mathbb{R}^{2}\) are going to carry over to \(\mathbb{C}^{2}\). From now on, we will denote the elements of \(\mathbb{C}^{2}\) as vectors. Vectors will be denoted in bold font.

Column Vectors

A typical convention while denoting vectors is to treat them as column vectors. Let us take an example:

\[ \begin{equation*} z =\begin{bmatrix} 1+i\\ 3-2i \end{bmatrix} \end{equation*} \] The vector \(\mathbf{z} \in \mathbb{C}^{2}\) and is treated as a \(2\times 1\) matrix. Note that we would have represented this as \(( 1+i,3-2i)\) while viewing \(\mathbb{C}^{2}\) as a simple set. Once we get the hang of column vectors, then the corresponding row vector would be:

\[ \begin{equation*} \mathbf{z}^{T} =\begin{bmatrix} 1+i & 3-2i \end{bmatrix} \end{equation*} \]

Addition

We can add vectors component wise. As an example, let \(\mathbf{z_{1}}\) and \(\mathbf{z_{2}}\) be two vectors in \(\mathbb{C}^{2}\):

\[ \begin{equation*} \mathbf{z_{1}} =\begin{bmatrix} i\\ 1+i \end{bmatrix} ,\mathbf{z_{2}} =\begin{bmatrix} -1+i\\ 5-3i \end{bmatrix} \end{equation*} \] Then, \(\mathbf{z_{1}} +\mathbf{z_{2}}\) is:

\[ \begin{equation*} \mathbf{z_{1}} +\mathbf{z_{2}} =\begin{bmatrix} -1+2i\\ 6-2i \end{bmatrix} \end{equation*} \]

Scalar multiplication

We can multiply a vector \(\mathbf{z}\) with a scalar. Note that the scalar can be any complex number. Let \(\alpha =2+i\) be a scalar and \(\mathbf{z} =\begin{bmatrix} 1\\ 3i \end{bmatrix}\) be a vector, then: \[ \begin{equation*} \alpha \mathbf{z} =\begin{bmatrix} 2+i\\ -3+6i \end{bmatrix} \end{equation*} \]

Conjugate

If \(\mathbf{z} =\begin{bmatrix} 2+i\\ -3+6i \end{bmatrix}\), then \(\mathbf{\overline{z}}\) is: \[ \begin{equation*} \mathbf{\overline{z}} =\begin{bmatrix} 2-i\\ -3-6i \end{bmatrix} \end{equation*} \] The conjugate is computed component-wise.

Conjugate transpose

This is not an entirely new operation but just a composition of two operations: conjugate and transpose. Conjugate and transpose are interchangeable operations. We could first take the transpose and then the conjugate or take the conjugate first and then the transpose. That is:

\[ \begin{equation*} (\mathbf{\overline{z}})^{T} =\overline{\mathbf{z}^{T}} \end{equation*} \] We will denote this by \(\overline{\mathbf{z}}^{T}\). As an example, if \(\mathbf{z} =\begin{bmatrix} 2+i\\ -3+6i \end{bmatrix}\), then \(\mathbf{\overline{z}}^{T}\) is: \[ \begin{equation*} \mathbf{\overline{z}}^{T} =\begin{bmatrix} 2-i & -3-6i \end{bmatrix} \end{equation*} \]

Linear combination

If we have three vectors \(\mathbf{z_{1}} ,\mathbf{z_{2}} ,\mathbf{z_{3}}\), then the following is a linear combination:

\[ \begin{equation*} ( 3+i)\mathbf{z_{1}} -( 5i)\mathbf{z_{2}} +( 2-6i)\mathbf{z_{3}} \end{equation*} \] This can be extended to a finite number of vectors:

\[ \begin{equation*} \alpha _{1}\mathbf{z_{1}} +\cdots +\alpha _{n}\mathbf{z_{n}} \end{equation*} \] Here, \(\mathbf{z_{i}} \in \mathbb{C}^{2}\) and \(\alpha _{i} \in \mathbb{C}\).

Inner Products

In addition to the operations of vector addition and scalar multiplication, we can add more structure to a vector space by defining the operation of an inner product. Recall that this results in what is called an inner product space. An inner product accepts two vectors as input and returns a scalar as output. In the case of complex vector spaces, recall that the scalar can be a complex number. In the case of \(\mathbb{R}^{2}\), we had the simple dot product as an inner product which returned a real number. Let \(\mathbf{x} ,\mathbf{y} \in \mathbb{R}^{2}\), then:

\[ \begin{equation*} \mathbf{x} =\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix} ,\ \mathbf{y} =\begin{bmatrix} y_{1}\\ y_{2} \end{bmatrix} \end{equation*} \] The dot product is

\[ \begin{equation*} \mathbf{x}^{T}\mathbf{y} =x_{1} y_{1} +x_{2} y_{2} \end{equation*} \] In general, an inner product is denoted using angle brackets \(\langle \ ,\ \rangle\). In the above case:

\[ \begin{equation*} \langle \mathbf{x} ,\mathbf{y} \rangle =\mathbf{x}^{T}\mathbf{y} \end{equation*} \] Note that the idea of an inner product is more abstract than the dot product. As a simple analogy consider the astronomical terms “sun” and “star”. The term “star” can be identified with the inner product, while the term “sun” can be identified with the dot product.

Why do we need inner products? An inner product introduces the geometrical ideas of lengths and angles into vector spaces. If we just have a vector space, all that we can do is add vectors and multiply them with scalars. It is only with the help of the inner product that we can talk about notions such as lengths, distances and angles.

Let us pause for a moment and consider \(\mathbb{R}^{2}\). If \(\mathbf{x} \in \mathbb{R}^{2}\), then we can define the norm of \(\mathbf{x}\) as \(||\mathbf{x} ||=\sqrt{\mathbf{x}^{T}\mathbf{x}} =\sqrt{x_{1}^{2} +x_{2}^{2}}\). The norm of \(\mathbf{x}\) gives us the notion of the length of a vector. For instance, if \(\mathbf{x} =\begin{bmatrix} 3\\ 4 \end{bmatrix}\), then \(\sqrt{\mathbf{x}^{T}\mathbf{x}} =5\) and is the length of the vector \(\mathbf{x}\). We would like something similar to hold for complex inner products.

Complex Inner Product

Coming back to \(\mathbb{C}^{2}\), is the dot product a valid inner product? Let us try. Consider the vector \(\mathbf{z} =\begin{bmatrix} 1\\ i \end{bmatrix}\). Now, let us compute \(\mathbf{z}^{T}\mathbf{z}\). \[ \begin{equation*} \begin{aligned} \mathbf{z}^{T}\mathbf{z} & =\begin{bmatrix} 1 & i \end{bmatrix}\begin{bmatrix} 1\\ i \end{bmatrix}\\ & =1+i^{2}\\ & =0 \end{aligned} \end{equation*} \] We wouldn’t want the “length” of a non-zero vector to be zero. Hence, the dot product doesn’t seem like a valid inner product for the complex vector space \(\mathbb{C}^{2}\). So what is a valid inner product for \(\mathbb{C}^{2}\)? The following is the standard inner product that we have for complex vector spaces:

\[ \begin{equation*} \langle \mathbf{x} ,\mathbf{y} \rangle =\overline{\mathbf{x}}^{T}\mathbf{y} \end{equation*} \] Here, \(\mathbf{x} ,\mathbf{y} \in \mathbb{C}^{2}\). Let us denote \(\mathbf{x} =\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}\) and \(\mathbf{y} =\begin{bmatrix} y_{1}\\ y_{2} \end{bmatrix}\) and understand what this means. Note that \(x_{1} ,x_{2} ,y_{1} ,y_{2} \in \mathbb{C}\): \[ \begin{equation*} \begin{aligned} \overline{\mathbf{x}}^{T}\mathbf{y} & =\begin{bmatrix} \overline{x_{1}} & \overline{x_{2}} \end{bmatrix}\begin{bmatrix} y_{1}\\ y_{2} \end{bmatrix}\\ & \\ & =\overline{x_{1}} y_{1} +\overline{x_{2}} y_{2} \end{aligned} \end{equation*} \] Let us now go back to the example \(z=\begin{bmatrix} 1\\ i \end{bmatrix}\) and see what this inner product does when both the vectors are \(\mathbf{z}\): \[ \begin{gather*} \begin{aligned} \overline{\mathbf{z}}^{T}\mathbf{z} & =\begin{bmatrix} 1 & -i \end{bmatrix}\begin{bmatrix} 1\\ i \end{bmatrix}\\ & =1-i^{2}\\ & =2 \end{aligned}\\ \end{gather*} \] This is much more reasonable than the simple dot product. For want of a better term, let us call the standard inner product for complex vector spaces the “complex inner product”. Let us now do an example: \[ \begin{equation*} \mathbf{x} =\begin{bmatrix} 1+i\\ 3-i \end{bmatrix} ,\mathbf{y} =\begin{bmatrix} 2i\\ i \end{bmatrix} \end{equation*} \] Then:

\[ \begin{equation*} \begin{aligned} \overline{\mathbf{x}}^{T}\mathbf{y} & =\begin{bmatrix} 1-i & 3+i \end{bmatrix}\begin{bmatrix} 2i\\ i \end{bmatrix}\\ & \\ & =( 1-i)( 2i) +( 3+i) i\\ & \\ & =2i-2i^{2} +3i+i^{2}\\ & \\ & =1+5i \end{aligned} \end{equation*} \] Note again that the inner product is a scalar and in the case of complex vector spaces, it can be any complex number.

Properties

Unlike the dot product for real vector spaces, the complex inner product is not symmetric. That is:

\[ \begin{equation*} \mathbf{\overline{x}}^{T}\mathbf{y} \neq \mathbf{\overline{y}}^{T}\mathbf{x} \end{equation*} \] Let us see this with an example:

\[ \begin{equation*} \mathbf{x} =\begin{bmatrix} 1-i\\ 2+i \end{bmatrix} ,y=\begin{bmatrix} 2-i\\ i \end{bmatrix} \end{equation*} \] Then, \[ \begin{equation*} \begin{aligned} \overline{\mathbf{x}}^{T}\mathbf{y} & =\begin{bmatrix} 1+i & 2-i \end{bmatrix}\begin{bmatrix} 2-i\\ i \end{bmatrix}\\ & =( 1+i)( 2-i) +( 2-i) i\\ & =2-i+2i-i^{2} +2i-i^{2}\\ & =4+3i \end{aligned} \end{equation*} \] And, \[ \begin{equation*} \begin{aligned} \overline{\mathbf{y}}^{T}\mathbf{x} & =\begin{bmatrix} 2+i & -i \end{bmatrix}\begin{bmatrix} 1-i\\ 2+i \end{bmatrix}\\ & =( 2+i)( 1-i) -i( 2+i)\\ & =2-2i+i-i^{2} -2i-i^{2}\\ & =4-3i \end{aligned} \end{equation*} \] We see that \(\overline{\mathbf{x}}^{T}\mathbf{y} \neq \overline{\mathbf{y}}^{T}\mathbf{x}\). But interestingly, we see that \(\overline{\overline{\mathbf{x}}^{T}\mathbf{y}} =\overline{\mathbf{y}}^{T}\mathbf{x}\). Though this inner product is not symmetric, it is “conjugate symmetric”. This will be the first property:

Conjugate Symmetry

If \(\mathbf{x}\) and \(\mathbf{y}\) are two vectors in \(\mathbb{C}^{2}\), then: \[ \begin{equation*} \overline{\overline{\mathbf{x}}^{T}\mathbf{y}} =\overline{\mathbf{y}}^{T}\mathbf{x} \end{equation*} \] Next, we have linearity.

Linearity

If \(\mathbf{x} ,\mathbf{y} ,\mathbf{z} \in \mathbb{C}^{2}\) and \(\alpha ,\beta \in \mathbb{C}\), then: \[ \begin{equation*} \overline{\mathbf{x}}^{T}(\mathbf{y} +\mathbf{z}) =\overline{\mathbf{x}}^{T}\mathbf{y} +\overline{\mathbf{x}}^{T}\mathbf{z} \end{equation*} \]

\[ \begin{equation*} \overline{\mathbf{x}}^{T}( \alpha \mathbf{y}) =\alpha \overline{x}^{T}\mathbf{y} \end{equation*} \]

The complex inner product is linear in the second argument. Rather than scaling the second argument, if we scale the first argument, we get the following:

\[ \begin{equation*} \overline{( \alpha \mathbf{x})}^{T}\mathbf{y} =\overline{\alpha }\left(\overline{x}^{T}\mathbf{y}\right) \end{equation*} \] Finally, we have the following property:

Positive-definiteness

If \(\mathbf{x}\) is any non-zero vector in \(\mathbb{C}^{2}\), then: \[ \begin{equation*} \mathbf{\overline{x}}^{T}\mathbf{x} >0 \end{equation*} \] There is more to this statement than meets the eye. The inner product returns a complex number. Complex numbers do not admit comparisons. For instance, a statement like \(1+2i >0\) makes no sense. However, the above property is making some such comparison. This is allowed because, \(\overline{\mathbf{x}}^{T}\mathbf{x}\) is always going to be a positive real number. Let us verify this with \(\mathbf{x} =\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}\): \[ \begin{gather*} \begin{aligned} \overline{\mathbf{x}}^{T}\mathbf{x} & =\overline{x_{1}} x_{1} +\overline{x_{2}} x_{2}\\ & =|x_{1} |^{2} +|x_{2} |^{2}\\ & >0 \end{aligned}\\ \end{gather*} \] Note that \(\overline{\mathbf{x}}^{T}\mathbf{x} =0\) if and only if \(\mathbf{x} =\mathbf{0}\). In other words, the complex inner product of a vector with itself is zero if and only if the vector itself is the zero vector (\(\mathbf{0}\)).

One frequent use of the inner product is to check if two vectors are orthogonal (perpendicular). We say that two vectors \(\mathbf{x}\) and \(\mathbf{y}\) are orthogonal (or \(\mathbf{x} \perp \mathbf{y}\)) if \(\overline{\mathbf{x}}^{T}\mathbf{y} =0\).

Notation Revisited

So far we have been using \(\overline{\mathbf{x}}^{T}\) to denote the conjugate transpose. This notation could be a bit cumbersome. There is an alternative notation that we will start using from now:

\[ \begin{equation*} \mathbf{x}^{H} \ \text{or} \ \mathbf{x}^{*} \end{equation*} \] \(H\) stands for Hermitian. Using this notation, the complex inner product \(\overline{\mathbf{x}}^{T}\mathbf{y}\) can be written as:

\[ \begin{equation*} \mathbf{x}^{*}\mathbf{y} \ \ \text{or} \ \ \mathbf{x}^{H}\mathbf{y} \end{equation*} \]

This can be read as “x star y” and “x Hermitian y” respectively.