Ferkans — Interactive Telecom Tutor

Why Matrices Are Central to Telecommunications

Every linear operation in a wireless communication system can be represented as a matrix acting on a vector. The MIMO channel relates the transmit signal vector $\mathbf{x} \in \mathbb{C}^{n_t}$ to the received signal vector $\mathbf{y} \in \mathbb{C}^{n_r}$ via

$\mathbf{y} = \mathbf{H}\mathbf{x} + \mathbf{n},$

where $\mathbf{H} \in \mathbb{C}^{n_r \times n_t}$ is the channel matrix and $\mathbf{n}$ is additive noise. A precoder at the transmitter is a matrix $\mathbf{F}$ applied before transmission; a combiner at the receiver is a matrix $\mathbf{W}^H$ applied after reception. The effective input–output relation becomes $\tilde{\mathbf{y}} = \mathbf{W}^H \mathbf{H} \mathbf{F} \mathbf{s} + \mathbf{W}^H \mathbf{n}$ .

Understanding the algebraic properties of these matrices — their rank, range, null space, and special structure (Hermitian, unitary, positive definite) — is therefore not an abstract exercise but the foundation upon which transceiver design, capacity analysis, and signal processing algorithms rest.

Definition:
Linear Map

Let $V$ and $W$ be vector spaces over the same field $\mathbb{F}$ (typically $\mathbb{R}$ or $\mathbb{C}$ ). A function $T : V \to W$ is a linear map (or linear transformation) if for all $\mathbf{x}, \mathbf{y} \in V$ and all $\alpha \in \mathbb{F}$ :

Additivity: $T(\mathbf{x} + \mathbf{y}) = T(\mathbf{x}) + T(\mathbf{y})$ .
Homogeneity: $T(\alpha \mathbf{x}) = \alpha\, T(\mathbf{x})$ .

Equivalently, $T$ is linear if and only if $T(\alpha \mathbf{x} + \beta \mathbf{y}) = \alpha\, T(\mathbf{x}) + \beta\, T(\mathbf{y})$ for all $\alpha, \beta \in \mathbb{F}$ and $\mathbf{x}, \mathbf{y} \in V$ .

Once bases are chosen for $V$ and $W$ , every linear map $T$ is represented uniquely by a matrix $\mathbf{A} \in \mathbb{F}^{m \times n}$ (where $n = \dim V$ , $m = \dim W$ ) such that $T(\mathbf{x}) = \mathbf{A}\mathbf{x}$ . Conversely, every matrix defines a linear map. This correspondence is an isomorphism of vector spaces: $\mathcal{L}(V, W) \cong \mathbb{F}^{m \times n}$ .

Definition:
Matrix–Vector and Matrix–Matrix Products

Let $\mathbf{A} = [a_{ij}] \in \mathbb{C}^{m \times n}$ and $\mathbf{B} = [b_{jk}] \in \mathbb{C}^{n \times p}$ .

Matrix–vector product. For $\mathbf{x} \in \mathbb{C}^n$ , the product $\mathbf{y} = \mathbf{A}\mathbf{x} \in \mathbb{C}^m$ has entries $y_i = \sum_{k=1}^{n} a_{ik}\, x_k, \qquad i = 1, \ldots, m.$ Equivalently, $\mathbf{A}\mathbf{x}$ is a linear combination of the columns of $\mathbf{A}$ : if $\mathbf{a}_1, \ldots, \mathbf{a}_n$ are the columns, then $\mathbf{A}\mathbf{x} = x_1 \mathbf{a}_1 + \cdots + x_n \mathbf{a}_n$ .

Matrix–matrix product. The product $\mathbf{C} = \mathbf{A}\mathbf{B} \in \mathbb{C}^{m \times p}$ has entries $c_{ik} = \sum_{j=1}^{n} a_{ij}\, b_{jk}, \qquad i = 1,\ldots,m,\; k = 1,\ldots,p.$ Column $k$ of $\mathbf{C}$ equals $\mathbf{A}\mathbf{b}_k$ , where $\mathbf{b}_k$ is column $k$ of $\mathbf{B}$ .

The column-view of the matrix–vector product is fundamental: it shows that the output $\mathbf{y} = \mathbf{A}\mathbf{x}$ always lies in the column space of $\mathbf{A}$ , no matter what $\mathbf{x}$ is. This observation is the key to understanding range and rank.

Definition:
Range (Column Space)

The range (or column space) of a matrix $\mathbf{A} \in \mathbb{C}^{m \times n}$ is $\mathcal{R}(\mathbf{A}) = \{\mathbf{A}\mathbf{x} : \mathbf{x} \in \mathbb{C}^n\} = \text{span}(\mathbf{a}_1, \ldots, \mathbf{a}_n) \subseteq \mathbb{C}^m,$ where $\mathbf{a}_1, \ldots, \mathbf{a}_n$ are the columns of $\mathbf{A}$ .

$\mathcal{R}(\mathbf{A})$ is a subspace of $\mathbb{C}^m$ . It is the image of the linear map $\mathbf{x} \mapsto \mathbf{A}\mathbf{x}$ : it tells you which output vectors are achievable. In a MIMO system, $\mathcal{R}(\mathbf{H})$ is the subspace of receive-signal space that the channel can excite.

Definition:
Null Space (Kernel)

The null space (or kernel) of a matrix $\mathbf{A} \in \mathbb{C}^{m \times n}$ is $\mathcal{N}(\mathbf{A}) = \{\mathbf{x} \in \mathbb{C}^n : \mathbf{A}\mathbf{x} = \mathbf{0}\} \subseteq \mathbb{C}^n.$

$\mathcal{N}(\mathbf{A})$ is a subspace of $\mathbb{C}^n$ (the domain, not the codomain). It consists of all inputs that the linear map "kills." A nonzero null space means the map is not injective. In MIMO, any transmit vector $\mathbf{x} \in \mathcal{N}(\mathbf{H})$ produces zero received signal — the channel is "blind" in those directions.

Definition:
Rank of a Matrix

The rank of $\mathbf{A} \in \mathbb{C}^{m \times n}$ is $\text{rank}(\mathbf{A}) = \dim\!\big(\mathcal{R}(\mathbf{A})\big).$ Equivalently, $\text{rank}(\mathbf{A})$ equals:

the maximum number of linearly independent columns of $\mathbf{A}$ , or
the maximum number of linearly independent rows of $\mathbf{A}$ (row rank equals column rank), or
the number of nonzero singular values of $\mathbf{A}$ .

Since $\text{rank}(\mathbf{A}) \leq \min(m, n)$ , a matrix is called full rank when equality holds. In MIMO, the rank of the channel matrix $\mathbf{H}$ determines the maximum number of independent data streams (spatial multiplexing gain) that can be transmitted.

Definition:
Invertibility

A square matrix $\mathbf{A} \in \mathbb{C}^{n \times n}$ is invertible (or nonsingular) if there exists a matrix $\mathbf{A}^{-1} \in \mathbb{C}^{n \times n}$ such that $\mathbf{A}\mathbf{A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}_n.$ The following are equivalent:

$\mathbf{A}$ is invertible.
$\text{rank}(\mathbf{A}) = n$ .
$\mathcal{N}(\mathbf{A}) = \{\mathbf{0}\}$ .
$\det(\mathbf{A}) \neq 0$ .
All eigenvalues of $\mathbf{A}$ are nonzero.

Invertibility means the linear map $\mathbf{x} \mapsto \mathbf{A}\mathbf{x}$ is a bijection: every output has a unique pre-image. When the channel matrix is square and invertible, zero-forcing detection amounts to multiplying the received signal by $\mathbf{H}^{-1}$ .

Definition:
Change of Basis

Let $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}$ and $\mathcal{B}' = \{\mathbf{b}'_1, \ldots, \mathbf{b}'_n\}$ be two ordered bases of a vector space $V$ . The change-of-basis matrix from $\mathcal{B}$ to $\mathcal{B}'$ is the unique invertible matrix $\mathbf{P} \in \mathbb{C}^{n \times n}$ whose $k$ -th column contains the coordinates of $\mathbf{b}_k$ in basis $\mathcal{B}'$ : $\mathbf{b}_k = \sum_{i=1}^{n} p_{ik}\, \mathbf{b}'_i, \qquad k = 1,\ldots,n.$ If $[\mathbf{x}]_{\mathcal{B}}$ denotes the coordinate vector of $\mathbf{x}$ in basis $\mathcal{B}$ , then $[\mathbf{x}]_{\mathcal{B}'} = \mathbf{P}^{-1} [\mathbf{x}]_{\mathcal{B}}.$ The matrix representation of a linear map $T$ transforms as $[\mathbf{A}]_{\mathcal{B}'} = \mathbf{P}^{-1} [\mathbf{A}]_{\mathcal{B}}\, \mathbf{P}.$

In wireless systems, change of basis is used constantly. For example, switching between the antenna-domain representation and the beamspace (angular) representation of a channel is a change of basis performed by the DFT matrix. Eigendecomposition and SVD can also be viewed as finding particularly revealing bases.

Definition:
Hermitian Matrix

A matrix $\mathbf{A} \in \mathbb{C}^{n \times n}$ is Hermitian (or self-adjoint) if $\mathbf{A}^H = \mathbf{A}$ , where $(\cdot)^H$ denotes the conjugate transpose.

Hermitian matrices are the complex generalization of real symmetric matrices. Every covariance matrix in wireless communications is Hermitian positive semidefinite.

Definition:
Unitary Matrix

A matrix $\mathbf{U} \in \mathbb{C}^{n \times n}$ is unitary if $\mathbf{U}^H \mathbf{U} = \mathbf{U} \mathbf{U}^H = \mathbf{I}_n$ .

Unitary matrices preserve inner products and norms: $\langle \mathbf{U}\mathbf{x}, \mathbf{U}\mathbf{y} \rangle = \langle \mathbf{x}, \mathbf{y} \rangle$ . In wireless, DFT matrices and beamforming codebooks are often unitary.

Definition:
Positive Definite and Positive Semidefinite Matrices

A Hermitian matrix $\mathbf{A} \in \mathbb{C}^{n \times n}$ is:

Positive definite ( $\mathbf{A} \succ 0$ ) if $\mathbf{x}^H \mathbf{A} \mathbf{x} > 0$ for all nonzero $\mathbf{x} \in \mathbb{C}^n$ .
Positive semidefinite ( $\mathbf{A} \succeq 0$ ) if $\mathbf{x}^H \mathbf{A} \mathbf{x} \geq 0$ for all $\mathbf{x} \in \mathbb{C}^n$ .

Every covariance matrix $\mathbf{R} = \mathbb{E}[\mathbf{x}\mathbf{x}^H]$ is positive semidefinite. The MIMO capacity formula $\log\det(\mathbf{I} + \text{SNR} \cdot \mathbf{H}\mathbf{H}^H)$ requires $\mathbf{H}\mathbf{H}^H \succeq 0$ .

Theorem: Rank–Nullity Theorem

For any matrix $\mathbf{A} \in \mathbb{C}^{m \times n}$ : $\text{rank}(\mathbf{A}) + \dim(\mathcal{N}(\mathbf{A})) = n.$

The domain $\mathbb{C}^n$ splits into two complementary pieces: the directions that $\mathbf{A}$ maps to nonzero outputs (contributing to the rank) and the directions that $\mathbf{A}$ collapses to zero (the null space). Together they must account for every dimension of the domain. In a MIMO channel with $n_t$ transmit antennas, if the channel has rank $r$ , then $r$ independent streams survive, while $n_t - r$ transmit directions are "wasted" (they lie in the null space of $\mathbf{H}$ ).

Show Hint

Construct a basis for $\mathcal{N}(\mathbf{A})$ and extend it to a basis of $\mathbb{C}^n$ .

Show that $\mathbf{A}$ maps the extension vectors to linearly independent vectors in $\mathcal{R}(\mathbf{A})$ .

Conclude by counting dimensions.

Proof

Step 1: Set up notation

Let $r = \text{rank}(\mathbf{A})$ and let $k = \dim(\mathcal{N}(\mathbf{A}))$ . We must show $r + k = n$ .

Choose a basis $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ for $\mathcal{N}(\mathbf{A})$ . Since $\mathcal{N}(\mathbf{A})$ is a subspace of $\mathbb{C}^n$ , by the basis extension theorem we can extend this to a basis of $\mathbb{C}^n$ : $\{\mathbf{v}_1, \ldots, \mathbf{v}_k, \mathbf{w}_1, \ldots, \mathbf{w}_{n-k}\}.$

Step 2: Show the images of the extension vectors span $\mathcal{R}(\mathbf{A})$

We claim that $\{\mathbf{A}\mathbf{w}_1, \ldots, \mathbf{A}\mathbf{w}_{n-k}\}$ spans $\mathcal{R}(\mathbf{A})$ .

Take any $\mathbf{y} \in \mathcal{R}(\mathbf{A})$ . Then $\mathbf{y} = \mathbf{A}\mathbf{x}$ for some $\mathbf{x} \in \mathbb{C}^n$ . Since $\{\mathbf{v}_1, \ldots, \mathbf{v}_k, \mathbf{w}_1, \ldots, \mathbf{w}_{n-k}\}$ is a basis of $\mathbb{C}^n$ , we can write $\mathbf{x} = \sum_{i=1}^{k} \alpha_i \mathbf{v}_i + \sum_{j=1}^{n-k} \beta_j \mathbf{w}_j.$ Applying $\mathbf{A}$ : $\mathbf{y} = \mathbf{A}\mathbf{x} = \sum_{i=1}^{k} \alpha_i \underbrace{\mathbf{A}\mathbf{v}_i}_{= \mathbf{0}} + \sum_{j=1}^{n-k} \beta_j \mathbf{A}\mathbf{w}_j = \sum_{j=1}^{n-k} \beta_j \mathbf{A}\mathbf{w}_j.$ Thus every vector in $\mathcal{R}(\mathbf{A})$ is a linear combination of $\mathbf{A}\mathbf{w}_1, \ldots, \mathbf{A}\mathbf{w}_{n-k}$ , so these vectors span $\mathcal{R}(\mathbf{A})$ .

Step 3: Show the images of the extension vectors are linearly independent

Suppose $\sum_{j=1}^{n-k} \gamma_j \mathbf{A}\mathbf{w}_j = \mathbf{0}$ . Then $\mathbf{A}\!\left(\sum_{j=1}^{n-k} \gamma_j \mathbf{w}_j\right) = \mathbf{0}$ , which means $\sum_{j=1}^{n-k} \gamma_j \mathbf{w}_j \in \mathcal{N}(\mathbf{A})$ .

Since $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ is a basis of $\mathcal{N}(\mathbf{A})$ , there exist scalars $\delta_1, \ldots, \delta_k$ such that $\sum_{j=1}^{n-k} \gamma_j \mathbf{w}_j = \sum_{i=1}^{k} \delta_i \mathbf{v}_i.$ Rearranging: $\sum_{i=1}^{k} (-\delta_i) \mathbf{v}_i + \sum_{j=1}^{n-k} \gamma_j \mathbf{w}_j = \mathbf{0}.$ But $\{\mathbf{v}_1, \ldots, \mathbf{v}_k, \mathbf{w}_1, \ldots, \mathbf{w}_{n-k}\}$ is a basis of $\mathbb{C}^n$ , hence linearly independent. Therefore all coefficients must be zero: $\delta_i = 0$ for all $i$ and $\gamma_j = 0$ for all $j$ .

This proves $\{\mathbf{A}\mathbf{w}_1, \ldots, \mathbf{A}\mathbf{w}_{n-k}\}$ is linearly independent.

Step 4: Conclude

From Steps 2 and 3, $\{\mathbf{A}\mathbf{w}_1, \ldots, \mathbf{A}\mathbf{w}_{n-k}\}$ is a basis of $\mathcal{R}(\mathbf{A})$ . Therefore: $r = \text{rank}(\mathbf{A}) = \dim(\mathcal{R}(\mathbf{A})) = n - k = n - \dim(\mathcal{N}(\mathbf{A})).$ Rearranging: $\text{rank}(\mathbf{A}) + \dim(\mathcal{N}(\mathbf{A})) = n$ . $\blacksquare$

Theorem: Properties of Hermitian Matrices

Let $\mathbf{A} \in \mathbb{C}^{n \times n}$ be Hermitian. Then:

All eigenvalues of $\mathbf{A}$ are real.
Eigenvectors corresponding to distinct eigenvalues are orthogonal.
The diagonal entries $a_{ii}$ are real for all $i$ .

A Hermitian matrix is "the same as its own adjoint," so it interacts symmetrically with inner products. This forces eigenvalues to be real (no imaginary component can survive the symmetry) and eigenvectors from different eigenvalues to be orthogonal.

Show Hint

For (1): compute $\lambda \langle \mathbf{x}, \mathbf{x} \rangle = \langle \mathbf{x}, \mathbf{A}\mathbf{x} \rangle$ and use $\mathbf{A}^H = \mathbf{A}$ .

For (2): compute $\lambda_1 \langle \mathbf{x}_2, \mathbf{x}_1 \rangle$ two different ways.

Proof

Proof of (1): Eigenvalues are real

Let $\mathbf{A}\mathbf{x} = \lambda \mathbf{x}$ with $\mathbf{x} \neq \mathbf{0}$ . Compute the inner product $\langle \mathbf{x}, \mathbf{A}\mathbf{x} \rangle$ : $\langle \mathbf{x}, \mathbf{A}\mathbf{x} \rangle = \mathbf{x}^H \mathbf{A} \mathbf{x} = \mathbf{x}^H (\lambda \mathbf{x}) = \lambda \|\mathbf{x}\|^2.$ On the other hand, since $\mathbf{A}^H = \mathbf{A}$ : $\langle \mathbf{x}, \mathbf{A}\mathbf{x} \rangle = (\mathbf{A}^H \mathbf{x})^H \mathbf{x} = (\mathbf{A}\mathbf{x})^H \mathbf{x} = (\lambda \mathbf{x})^H \mathbf{x} = \bar{\lambda} \|\mathbf{x}\|^2.$ Since $\|\mathbf{x}\|^2 > 0$ , we conclude $\lambda = \bar{\lambda}$ , so $\lambda \in \mathbb{R}$ . $\blacksquare$

Proof of (2): Orthogonality of eigenvectors

Let $\mathbf{A}\mathbf{x}_1 = \lambda_1 \mathbf{x}_1$ and $\mathbf{A}\mathbf{x}_2 = \lambda_2 \mathbf{x}_2$ with $\lambda_1 \neq \lambda_2$ . $\lambda_1 \langle \mathbf{x}_2, \mathbf{x}_1 \rangle = \langle \mathbf{x}_2, \mathbf{A}\mathbf{x}_1 \rangle = \langle \mathbf{A}^H \mathbf{x}_2, \mathbf{x}_1 \rangle = \langle \mathbf{A}\mathbf{x}_2, \mathbf{x}_1 \rangle = \overline{\lambda_2} \langle \mathbf{x}_2, \mathbf{x}_1 \rangle = \lambda_2 \langle \mathbf{x}_2, \mathbf{x}_1 \rangle,$ where the last step uses $\lambda_2 \in \mathbb{R}$ from part (1). Thus $(\lambda_1 - \lambda_2)\langle \mathbf{x}_2, \mathbf{x}_1 \rangle = 0$ . Since $\lambda_1 \neq \lambda_2$ , we must have $\langle \mathbf{x}_2, \mathbf{x}_1 \rangle = 0$ . $\blacksquare$

Proof of (3): Real diagonal entries

By definition, $\mathbf{A}^H = \mathbf{A}$ means $\bar{a}_{ji} = a_{ij}$ for all $i, j$ . Setting $i = j$ : $\bar{a}_{ii} = a_{ii}$ , so each diagonal entry is real. $\blacksquare$

Theorem: Unitary Matrices Preserve Inner Products and Norms

Let $\mathbf{U} \in \mathbb{C}^{n \times n}$ be unitary. Then for all $\mathbf{x}, \mathbf{y} \in \mathbb{C}^n$ :

$\langle \mathbf{U}\mathbf{x}, \mathbf{U}\mathbf{y} \rangle = \langle \mathbf{x}, \mathbf{y} \rangle$ (inner product preservation).
$\|\mathbf{U}\mathbf{x}\| = \|\mathbf{x}\|$ (norm / energy preservation).

Unitary transformations are "rigid rotations" (possibly with reflections) of complex space: they rotate and relabel axes without stretching or compressing any direction. This is why they preserve lengths and angles.

Show Hint

Expand $\langle \mathbf{U}\mathbf{x}, \mathbf{U}\mathbf{y} \rangle = (\mathbf{U}\mathbf{x})^H (\mathbf{U}\mathbf{y})$ .

Proof

$\langle \mathbf{U}\mathbf{x}, \mathbf{U}\mathbf{y} \rangle = (\mathbf{U}\mathbf{x})^H (\mathbf{U}\mathbf{y}) = \mathbf{x}^H \mathbf{U}^H \mathbf{U} \mathbf{y} = \mathbf{x}^H \mathbf{I}_n \mathbf{y} = \mathbf{x}^H \mathbf{y} = \langle \mathbf{x}, \mathbf{y} \rangle.$ $Setting$ \mathbf{y} = \mathbf{x} $:$ |\mathbf{U}\mathbf{x}|^2 = \langle \mathbf{U}\mathbf{x}, \mathbf{U}\mathbf{x} \rangle = \langle \mathbf{x}, \mathbf{x} \rangle = |\mathbf{x}|^2 $, hence$ |\mathbf{U}\mathbf{x}| = |\mathbf{x}| $.$ \blacksquare$

Special Matrix Classes

Property	Hermitian ( $\mathbf{A}^H = \mathbf{A}$ )	Unitary ( $\mathbf{U}^H\mathbf{U} = \mathbf{I}$ )	Positive Definite ( $\mathbf{A} \succ 0$ )
Eigenvalues	Real	On unit circle ( $\|\lambda\| = 1$ )	Real and positive
Diagonalizable?	Always (spectral theorem)	Always	Always (since Hermitian)
Determinant	Real	$\|\det\| = 1$	Real and positive
Inverse	$\mathbf{A}^{-1}$ is Hermitian (if it exists)	$\mathbf{U}^{-1} = \mathbf{U}^H$ (always exists)	$\mathbf{A}^{-1}$ is also positive definite
Preserves norms?	Not in general	Yes: $\\|\mathbf{U}\mathbf{x}\\| = \\|\mathbf{x}\\|$	Not in general
Quadratic form $\mathbf{x}^H \mathbf{A} \mathbf{x}$	Always real	Complex in general	Always real and positive (for $\mathbf{x} \neq \mathbf{0}$ )
Telecom example	Covariance matrix $\mathbf{R}_{\mathbf{x}}$	DFT matrix, beamforming codebook	Noise covariance $\mathbf{R}_{\mathbf{n}} \succ 0$
Requires Hermitian?	By definition	No	Yes, by definition
Closure under addition?	Yes	No	Yes (cone)
Closure under multiplication?	No (product of two Hermitian matrices is Hermitian iff they commute)	Yes	No in general

Example: Computing Rank, Range, and Null Space

Let $\mathbf{A} = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 3 & 6 & 3 \end{bmatrix} \in \mathbb{R}^{3 \times 3}.$ Find $\text{rank}(\mathbf{A})$ , $\mathcal{R}(\mathbf{A})$ , and $\mathcal{N}(\mathbf{A})$ . Verify the rank–nullity theorem.

Solution

Step 1: Row reduce $\mathbf{A}$

Apply elementary row operations: $\mathbf{A} = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 3 & 6 & 3 \end{bmatrix} \xrightarrow{R_2 \leftarrow R_2 - 2R_1} \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 3 & 6 & 3 \end{bmatrix} \xrightarrow{R_3 \leftarrow R_3 - 3R_1} \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.$ The row echelon form has one pivot (in column 1).

Step 2: Read off rank

The number of pivots is 1, so $\text{rank}(\mathbf{A}) = 1$ .

Step 3: Find the range $\mathcal{R}(\mathbf{A})$

The pivot is in column 1, so column 1 of the original matrix forms a basis for the column space: $\mathcal{R}(\mathbf{A}) = \text{span}\!\left(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\right).$ Indeed, column 2 is $2 \times$ column 1, and column 3 equals column 1.

Step 4: Find the null space $\mathcal{N}(\mathbf{A})$

Solve $\mathbf{A}\mathbf{x} = \mathbf{0}$ using the row echelon form. The system reduces to $x_1 + 2x_2 + x_3 = 0$ , with $x_2$ and $x_3$ as free variables. Setting $(x_2, x_3) = (1, 0)$ gives $x_1 = -2$ ; setting $(x_2, x_3) = (0, 1)$ gives $x_1 = -1$ . Thus: $\mathcal{N}(\mathbf{A}) = \text{span}\!\left( \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix},\; \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \right).$

Step 5: Verify rank–nullity

$\text{rank}(\mathbf{A}) + \dim(\mathcal{N}(\mathbf{A})) = 1 + 2 = 3 = n$ . $\checkmark$

Example: The $2 \times 2$ DFT Matrix Is Unitary

Let $\omega = e^{-j 2\pi / 2} = e^{-j\pi} = -1$ . The $2 \times 2$ DFT matrix is $\mathbf{F}_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & \omega \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}.$ Verify that $\mathbf{F}_2$ is unitary.

Solution

Compute $\mathbf{F}_2^H \mathbf{F}_2$

Since $\mathbf{F}_2$ is real in this case, $\mathbf{F}_2^H = \mathbf{F}_2^T$ . $\mathbf{F}_2^H \mathbf{F}_2 = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1+1 & 1-1 \\ 1-1 & 1+1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \mathbf{I}_2.$ Since $\mathbf{F}_2^H \mathbf{F}_2 = \mathbf{I}_2$ , the matrix is unitary. (For a square matrix, $\mathbf{U}^H \mathbf{U} = \mathbf{I}$ implies $\mathbf{U}\mathbf{U}^H = \mathbf{I}$ as well.)

Example: Checking Positive Definiteness

Show that the matrix $\mathbf{A} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ is positive definite.

Solution

Method 1: Eigenvalue test

The characteristic polynomial is $\det(\mathbf{A} - \lambda \mathbf{I}) = (2 - \lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3)$ . The eigenvalues are $\lambda_1 = 1 > 0$ and $\lambda_2 = 3 > 0$ . Since $\mathbf{A}$ is symmetric (hence Hermitian over $\mathbb{R}$ ) and all eigenvalues are strictly positive, $\mathbf{A} \succ 0$ .

Method 2: Direct quadratic form

For any $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \neq \mathbf{0}$ : $\mathbf{x}^T \mathbf{A} \mathbf{x} = 2x_1^2 + 2x_1 x_2 + 2x_2^2 = x_1^2 + (x_1 + x_2)^2 + x_2^2.$ This is a sum of squares, and at least one of $x_1, x_2$ is nonzero, so $x_1^2 + x_2^2 > 0$ . Adding the nonnegative term $(x_1 + x_2)^2$ , the entire expression is strictly positive. Thus $\mathbf{A} \succ 0$ .

Rank

The rank of a matrix $\mathbf{A} \in \mathbb{C}^{m \times n}$ is the dimension of its column space $\mathcal{R}(\mathbf{A})$ , equivalently the number of linearly independent columns (or rows). It satisfies $0 \leq \text{rank}(\mathbf{A}) \leq \min(m, n)$ .

Null Space (Kernel)

The null space of $\mathbf{A} \in \mathbb{C}^{m \times n}$ is the set of all vectors $\mathbf{x} \in \mathbb{C}^n$ satisfying $\mathbf{A}\mathbf{x} = \mathbf{0}$ . It is a subspace of $\mathbb{C}^n$ with dimension $n - \text{rank}(\mathbf{A})$ .

Hermitian Matrix

A square matrix $\mathbf{A}$ satisfying $\mathbf{A}^H = \mathbf{A}$ . Hermitian matrices have real eigenvalues, orthogonal eigenvectors, and are diagonalizable by a unitary similarity transformation (spectral theorem).

Unitary Matrix

A square matrix $\mathbf{U}$ satisfying $\mathbf{U}^H \mathbf{U} = \mathbf{I}$ . Unitary matrices preserve inner products and norms, have eigenvalues on the unit circle, and their inverse is simply $\mathbf{U}^H$ .

Positive Definite Matrix

A Hermitian matrix $\mathbf{A}$ such that $\mathbf{x}^H \mathbf{A} \mathbf{x} > 0$ for every nonzero vector $\mathbf{x}$ . Equivalently, all eigenvalues are strictly positive. Positive semidefinite relaxes " $> 0$ " to " $\geq 0$ ."

Quick Check

Let $\mathbf{A} \in \mathbb{C}^{4 \times 6}$ with $\text{rank}(\mathbf{A}) = 3$ . What is $\dim(\mathcal{N}(\mathbf{A}))$ ?

$1$

$3$

$4$

$6$

Correction:

3

By the rank–nullity theorem, $\dim(\mathcal{N}(\mathbf{A})) = n - \text{rank}(\mathbf{A}) = 6 - 3 = 3$ .

Quick Check

A matrix $\mathbf{H} \in \mathbb{C}^{8 \times 4}$ represents a MIMO channel with 4 transmit and 8 receive antennas. If $\mathcal{N}(\mathbf{H}) = \{\mathbf{0}\}$ , what is $\text{rank}(\mathbf{H})$ ?

$4$

$8$

$12$

Cannot be determined

Correction:

4

$\dim(\mathcal{N}(\mathbf{H})) = 0$ and $n = 4$ , so $\text{rank}(\mathbf{H}) = 4 - 0 = 4$ . The channel is full column rank, meaning all 4 transmit streams can be recovered.

Quick Check

Which of the following is always true for a unitary matrix $\mathbf{U}$ ?

$\mathbf{U}$ is Hermitian

$\det(\mathbf{U}) = 1$

$|\det(\mathbf{U})| = 1$

$\mathbf{U}$ is positive definite

Correction:

|\det(\mathbf{U})| = 1

From $\mathbf{U}^H \mathbf{U} = \mathbf{I}$ , taking determinants: $\overline{\det(\mathbf{U})} \cdot \det(\mathbf{U}) = 1$ , so $|\det(\mathbf{U})|^2 = 1$ , giving $|\det(\mathbf{U})| = 1$ . The determinant itself can be any complex number on the unit circle (e.g., $\det = -1$ for reflection matrices), so $\det(\mathbf{U}) = 1$ is not always true.

Quick Check

Let $\mathbf{A}$ be Hermitian. Which statement is false?

All eigenvalues of $\mathbf{A}$ are real.

All diagonal entries of $\mathbf{A}$ are real.

All entries of $\mathbf{A}$ are real.

$\mathbf{A}$ is diagonalizable.

Correction:

All entries of

\mathbf{A}

are real.

A Hermitian matrix satisfies $\bar{a}_{ji} = a_{ij}$ . The off-diagonal entries need not be real; they satisfy $a_{ji} = \bar{a}_{ij}$ (conjugate symmetry). For example, $\begin{bmatrix} 1 & j \\ -j & 2 \end{bmatrix}$ is Hermitian but has complex off-diagonal entries.

Quick Check

If $\mathbf{A} \succ 0$ and $\mathbf{B} \succ 0$ , is $\mathbf{A} + \mathbf{B} \succ 0$ ?

Yes, always

Only if $\mathbf{A}$ and $\mathbf{B}$ commute

Not necessarily

Correction:

Yes, always

For any nonzero $\mathbf{x}$ : $\mathbf{x}^H(\mathbf{A} + \mathbf{B})\mathbf{x} = \mathbf{x}^H \mathbf{A} \mathbf{x} + \mathbf{x}^H \mathbf{B} \mathbf{x} > 0 + 0 = 0$ . The set of positive definite matrices forms an open convex cone.

Common Mistake: Transpose vs. Conjugate Transpose for Complex Matrices

Mistake:

"I need the adjoint of my channel matrix, so I'll just transpose it: $\mathbf{H}^T$ ."

Correction:

For complex matrices, the adjoint (Hermitian transpose) is the conjugate transpose $\mathbf{H}^H = \overline{\mathbf{H}}^T$ , not the plain transpose $\mathbf{H}^T$ . Using $\mathbf{H}^T$ instead of $\mathbf{H}^H$ will give wrong results whenever $\mathbf{H}$ has complex entries.

For example, let $\mathbf{h} = \begin{bmatrix} 1 \\ j \end{bmatrix}$ . Then $\|\mathbf{h}\|^2 = \mathbf{h}^H \mathbf{h} = 1 + 1 = 2$ , but $\mathbf{h}^T \mathbf{h} = 1 + j^2 = 1 - 1 = 0$ .

The transpose gives a bilinear form; the conjugate transpose gives a sesquilinear (inner product) form. Only the latter guarantees $\mathbf{x}^H \mathbf{x} \geq 0$ with equality iff $\mathbf{x} = \mathbf{0}$ .

Tip: Rule of thumb: in complex-valued signal processing and wireless communications, always use $(\cdot)^H$ unless you have a specific algebraic reason to use $(\cdot)^T$ (e.g., when working with the $\text{vec}$ operator or Kronecker products where the transpose is genuinely intended).

Common Mistake: Rank Is Not the Number of Nonzero Rows or Columns

Mistake:

"My matrix has 3 nonzero rows, so its rank is 3."

Correction:

Rank is the number of linearly independent rows (or columns), not the number of nonzero ones. For example, $\mathbf{A} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$ has 2 nonzero rows but $\text{rank}(\mathbf{A}) = 1$ because row 2 is $2 \times$ row 1. Always row-reduce (or compute the SVD) to determine rank.

Tip: The most numerically reliable way to compute rank is via the SVD: count the number of singular values that exceed a suitable tolerance.

⚠️Engineering Note

Numerical Invertibility and Condition Numbers

The theoretical invertibility criterion $\det(\mathbf{A}) \neq 0$ is useless in floating-point arithmetic. A matrix with $\det(\mathbf{A}) = 10^{-15}$ is mathematically invertible but numerically singular. The practical criterion is the condition number $\kappa(\mathbf{A}) = \sigma_{\max}/\sigma_{\min}$ :

$\kappa < 10^3$ : well-conditioned, safe to invert.
$10^3 < \kappa < 10^{10}$ : ill-conditioned, use regularization (Tikhonov, diagonal loading).
$\kappa > 10^{10}$ (in 64-bit): effectively singular, do not invert. In MIMO detection, computing $\mathbf{H}^{-1}$ (zero-forcing) fails when the channel is ill-conditioned. Use MMSE regularization $(\mathbf{H}^H\mathbf{H} + \alpha\mathbf{I})^{-1}\mathbf{H}^H$ instead.

Practical Constraints

•
IEEE 754 double precision: ~15 significant digits, so matrices with $\kappa > 10^{15}$ lose all precision
•
32-bit (single precision): only ~7 digits; $\kappa > 10^7$ is dangerous
•
NumPy: np.linalg.cond(H) computes condition number; never use np.linalg.inv(H) without checking

🔧Engineering Note

Computing Rank in Practice: SVD Thresholding

The theoretical rank (number of nonzero singular values) is never exactly computable in floating-point. In practice, rank is determined by counting singular values above a tolerance: $\text{rank}_\epsilon(\mathbf{A}) = |\{i : \sigma_i > \epsilon\}|,$ where $\epsilon$ is typically $\epsilon = \max(m,n) \cdot \sigma_1 \cdot \varepsilon_{\text{mach}}$ (with $\varepsilon_{\text{mach}} \approx 2.2 \times 10^{-16}$ for double precision). This is the default used by numpy.linalg.matrix_rank.

Practical Constraints

•
SVD cost: $O(mn \min(m,n))$ for an $m \times n$ matrix
•
For large MIMO systems ( $n_t = 64$ ), the SVD of $\mathbf{H} \in \mathbb{C}^{64 \times 64}$ costs ~ $10^6$ flops

Key Takeaway

For $\mathbf{A} \in \mathbb{C}^{m \times n}$ , the domain $\mathbb{C}^n$ decomposes as $\mathbb{C}^n = \mathcal{R}(\mathbf{A}^H) \oplus \mathcal{N}(\mathbf{A})$ (an orthogonal direct sum). The rank–nullity theorem $\text{rank}(\mathbf{A}) + \dim(\mathcal{N}(\mathbf{A})) = n$ is the dimension count of this decomposition. In MIMO, this tells us exactly how many independent data streams the channel supports.

Key Takeaway

Three matrix classes appear everywhere in telecommunications:

Hermitian ( $\mathbf{A}^H = \mathbf{A}$ ): covariance matrices, Fisher information matrices, optimization cost matrices. Key property: real eigenvalues.
Unitary ( $\mathbf{U}^H \mathbf{U} = \mathbf{I}$ ): DFT matrices, beamforming codebooks, precoder/combiner matrices in capacity-achieving schemes. Key property: norm preservation.
Positive (semi)definite ( $\mathbf{A} \succeq 0$ ): covariance matrices, Gram matrices $\mathbf{H}^H \mathbf{H}$ . Key property: nonnegative eigenvalues guarantee meaningful power/energy interpretations.

The spectral theorem unifies them: every Hermitian matrix is unitarily diagonalizable with real eigenvalues, and it is positive (semi)definite iff those eigenvalues are all positive (nonneg.).

Key Takeaway

Many wireless signal processing operations are changes of basis in disguise. The DFT transforms from antenna domain to beamspace (angular domain). The SVD of a channel matrix provides the optimal transmit and receive bases that diagonalize the channel into independent parallel sub-channels. Recognizing an operation as a change of basis often reveals the underlying structure and suggests efficient implementations.

Why This Matters: The Channel Matrix Maps Transmit to Receive Signal Space

In a narrowband MIMO system with $n_t$ transmit antennas and $n_r$ receive antennas, the channel is described by $\mathbf{y} = \mathbf{H}\mathbf{x} + \mathbf{n},$ where $\mathbf{H} \in \mathbb{C}^{n_r \times n_t}$ is the channel matrix. The concepts of this section directly govern system performance:

Range: $\mathcal{R}(\mathbf{H}) \subseteq \mathbb{C}^{n_r}$ is the set of noiseless received signals achievable by varying $\mathbf{x}$ . Only the projection of $\mathbf{n}$ onto $\mathcal{R}(\mathbf{H})$ matters for detection; noise in $\mathcal{R}(\mathbf{H})^\perp$ is irrelevant.
Null space: $\mathcal{N}(\mathbf{H}) \subseteq \mathbb{C}^{n_t}$ consists of transmit directions that produce zero received signal. In multiuser MIMO, one user's precoder is deliberately chosen in $\mathcal{N}(\mathbf{H}_k)$ of the interfered user $k$ (zero-forcing).
Rank: $\text{rank}(\mathbf{H})$ is the number of independent spatial streams the channel supports. In rich scattering, $\text{rank}(\mathbf{H}) = \min(n_t, n_r)$ with probability 1 (full rank). In line-of-sight channels, $\text{rank}(\mathbf{H}) = 1$ (rank-deficient), and spatial multiplexing is impossible.
Hermitian structure: $\mathbf{H}^H \mathbf{H} \in \mathbb{C}^{n_t \times n_t}$ and $\mathbf{H}\mathbf{H}^H \in \mathbb{C}^{n_r \times n_r}$ are Hermitian positive semidefinite. Their eigenvalues are the squared singular values of $\mathbf{H}$ , which determine the SNR on each spatial sub-channel.
Unitary precoders/combiners: Capacity-achieving transmission on the MIMO channel uses the SVD $\mathbf{H} = \mathbf{U}\mathbf{\Sigma}\mathbf{V}^H$ . The transmit precoder $\mathbf{V}$ and receive combiner $\mathbf{U}^H$ are unitary, ensuring no energy is wasted in the transformation.

Why This Matters: Null Space and Zero-Forcing in Multiuser MIMO

Consider a base station with $n_t$ antennas serving two single-antenna users with channel vectors $\mathbf{h}_1^H, \mathbf{h}_2^H \in \mathbb{C}^{1 \times n_t}$ . To eliminate inter-user interference, the zero-forcing precoder for user 1 must satisfy $\mathbf{h}_2^H \mathbf{f}_1 = 0$ , i.e., $\mathbf{f}_1 \in \mathcal{N}(\mathbf{h}_2^H)$ .

By the rank–nullity theorem, $\dim(\mathcal{N}(\mathbf{h}_2^H)) = n_t - \text{rank}(\mathbf{h}_2^H) = n_t - 1$ (assuming $\mathbf{h}_2 \neq \mathbf{0}$ ). Thus the precoder has $n_t - 1$ degrees of freedom: enough to also maximize the signal power $|\mathbf{h}_1^H \mathbf{f}_1|^2$ subject to the null-space constraint. This is the foundation of zero-forcing beamforming, one of the most widely used multiuser MIMO techniques.

Matrices and Linear Maps

Why Matrices Are Central to Telecommunications

Definition: Linear Map

Definition: Matrix–Vector and Matrix–Matrix Products

Definition: Range (Column Space)

Definition: Null Space (Kernel)

Definition: Rank of a Matrix

Definition: Invertibility

Definition: Change of Basis

Definition: Hermitian Matrix

Definition: Unitary Matrix

Definition: Positive Definite and Positive Semidefinite Matrices

Theorem: Rank–Nullity Theorem

Step 1: Set up notation

Step 2: Show the images of the extension vectors span $\mathcal{R}(\mathbf{A})$

Step 3: Show the images of the extension vectors are linearly independent

Step 4: Conclude

Theorem: Properties of Hermitian Matrices

Proof of (1): Eigenvalues are real

Proof of (2): Orthogonality of eigenvectors

Proof of (3): Real diagonal entries

Theorem: Unitary Matrices Preserve Inner Products and Norms

Proof

Special Matrix Classes

Example: Computing Rank, Range, and Null Space

Step 1: Row reduce $\mathbf{A}$

Step 2: Read off rank

Step 3: Find the range $\mathcal{R}(\mathbf{A})$

Step 4: Find the null space $\mathcal{N}(\mathbf{A})$

Step 5: Verify rank–nullity

Example: The 2×22 \times 22×2 DFT Matrix Is Unitary

Compute $\mathbf{F}_2^H \mathbf{F}_2$

Example: Checking Positive Definiteness

Method 1: Eigenvalue test

Method 2: Direct quadratic form

Rank

Null Space (Kernel)

Hermitian Matrix

Unitary Matrix

Positive Definite Matrix

Quick Check

Quick Check

Quick Check

Quick Check

Quick Check

Common Mistake: Transpose vs. Conjugate Transpose for Complex Matrices

Common Mistake: Rank Is Not the Number of Nonzero Rows or Columns

Numerical Invertibility and Condition Numbers

Computing Rank in Practice: SVD Thresholding

Key Takeaway

Key Takeaway

Key Takeaway

Why This Matters: The Channel Matrix Maps Transmit to Receive Signal Space

Why This Matters: Null Space and Zero-Forcing in Multiuser MIMO

Definition:
Linear Map

Definition:
Matrix–Vector and Matrix–Matrix Products

Definition:
Range (Column Space)

Definition:
Null Space (Kernel)

Definition:
Rank of a Matrix

Definition:
Invertibility

Definition:
Change of Basis

Definition:
Hermitian Matrix

Definition:
Unitary Matrix

Definition:
Positive Definite and Positive Semidefinite Matrices

Example: The $2 \times 2$ DFT Matrix Is Unitary