Rank and nullity¶
Highlights: In this lecture we will show some most important results about the rank of a matrix:
- column rank = row rank.
- Matrix multiplication can only decrease the rank:
$\text{rank}(\mathbf{A} \mathbf{B}) \le \text{rank}(\mathbf{A})$
$\text{rank}(\mathbf{A} \mathbf{B}) \le \text{rank}(\mathbf{B})$.
Some special cases: (1) $\text{rank}(\mathbf{A} \mathbf{B}) = \text{rank}(\mathbf{A})$ if $\mathbf{B}$ has full row rank. (2) $\text{rank}(\mathbf{A} \mathbf{B}) = \text{rank}(\mathbf{B})$ if $\mathbf{A}$ has full column rank. (3). $\text{rank}(\mathbf{A}' \mathbf{A}) = \text{rank}(\mathbf{A})$ for any $\mathbf{A}$. - $\text{rank}(\mathbf{A} + \mathbf{B}) \le \text{rank}(\mathbf{B}) + \text{rank}(\mathbf{B})$.
- Rank-nullity theorem: For any $\mathbf{A} \in \mathbb{R}^{m \times n}$, $\text{rank}(\mathbf{A}) + \text{nullity}(\mathbf{A}) = n$.
- Fundamental theorem of ranks: $\text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}') = \text{rank}(\mathbf{A}'\mathbf{A}) = \text{rank}(\mathbf{A}\mathbf{A}')$.
using Pkg
Pkg.activate(pwd())
Pkg.instantiate()
Pkg.status()
using LinearAlgebra, RDatasets, StatsModels
# the famous Fisher's Iris data
# <https://en.wikipedia.org/wiki/Iris_flower_data_set>
iris = dataset("datasets", "iris")
# use full dummy coding (one-hot coding) for categorical variable Species
X = ModelMatrix(ModelFrame(
@formula(1 ~ 1 + SepalLength + SepalWidth + PetalLength + PetalWidth + Species),
iris,
contrasts = Dict(:Species => StatsModels.FullDummyCoding()))).m
@show size(X)
@show rank(X)
@show rank(X')
@show rank(X' * X)
@show rank(X * X');
# only one basis vector in N(X)
nullspace(X)
Rank¶
Let $\mathbf{A}$ be an $m \times n$ matrix \begin{eqnarray*} \mathbf{A} = \begin{pmatrix} \mid & & \mid \\ \mathbf{a}_1 & \ldots & \mathbf{a}_n \\ \mid & & \mid \end{pmatrix}. \end{eqnarray*}
The column rank of $\mathbf{A}$ is the maximum number of linearly independent columns of $\mathbf{A}$.
In other words, column rank of $\mathbf{A}$ is $\text{dim}(\mathcal{C}(\mathbf{A}))$.
The row rank of $\mathbf{A}$ is the maximum number of linearly independent rows of $\mathbf{A}$.
In other words, row rank of $\mathbf{A}$ is $\text{dim}(\mathcal{R}(\mathbf{A})) = \text{dim}(\mathcal{C}(\mathbf{A}'))$.
For any $m \times n$ matrix $\mathbf{A}$, its column rank is equal to the row rank, which we shall call the rank of $\mathbf{A}$.
We will give a simple proof using rank factorization below.
For any $\mathbf{A} \in \mathbb{R}^{m \times n}$, $\text{rank}(\mathbf{A}) \le \min \{m, n\}$.
A = [1 -2 -2; 3 -6 -6]
# column rank
rank(A)
# row rank
rank(A')
# another matrix
@show size(X)
@show rank(X)
For $\mathbf{A} \in \mathbb{R}^{m \times n}$, we say $\mathbf{A}$ is full rank if $\text{rank}(\mathbf{A}) = \min \{m, n\}$.
It is full row rank if $\text{rank}(\mathbf{A}) = m$.
It is full column rank if $\text{rank}(\mathbf{A}) = n$.
A square matrix $\mathbf{A} \in \mathbb{R}^{n \times n}$ is singular if $\text{rank}(\mathbf{A}) < n$ and non-singular or invertible if $\text{rank}(\mathbf{A}) = n$.
Example: The identity matrix $$ \mathbf{I} = \begin{pmatrix} 1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 1 \end{pmatrix} $$ is full rank.
Effects of matrix multiplication on rank¶
$\text{rank}(\mathbf{A}\mathbf{B}) \le \text{rank}(\mathbf{A})$ and $\text{rank}(\mathbf{A}\mathbf{B}) \le \text{rank}(\mathbf{B})$.
In words, matrix multiplication can only decrease the rank.
Proof: Because $\mathcal{C}(\mathbf{A}\mathbf{B}) \subseteq \mathcal{C}(\mathbf{A})$ (why?), we have $\text{rank}(\mathbf{A}\mathbf{B}) \le \text{rank}(\mathbf{A})$ by monotonicity of dimension. Similary, because the row space of $\mathbf{A}\mathbf{B}$ is a subset of the row space of $\mathbf{B}$, we have $\text{rank}(\mathbf{A}\mathbf{B}) \le \text{rank}(\mathbf{B})$.
$\text{rank}(\mathbf{A}\mathbf{B}) = \text{rank}(\mathbf{A})$ if $\mathbf{B}$ is square and of full rank. More general, left-multiplying by a matrix with full column rank or right-multiplying by a matrix of full row rank does not change rank.
Proof (optional): We show the more general statement. Assume $\mathbf{B} \in \mathbb{R}^{m \times n}$ has full row rank, we want to show $\text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}\mathbf{B})$. Since $\mathbf{B} \in \mathbb{R}^{m \times n}$ has full row rank, there exists a permutation matrix $\mathbf{P} \in \{0,1\}^{n \times n}$ such that $$ \mathbf{B} \mathbf{P} = \begin{pmatrix} \mathbf{B}_1 : \mathbf{B}_2 \end{pmatrix}, $$ where $\mathbf{B}_1 \in \mathbb{R}^{m \times m}$ is non-singular and $\mathbf{B}_2 \in \mathbb{R}^{m \times (n-m)}$. Then $$ \text{rank}(\mathbf{A}) \ge \text{rank}(\mathbf{A}\mathbf{B}) = \text{rank}(\mathbf{A} \begin{pmatrix} \mathbf{B}_1 : \mathbf{B}_2 \end{pmatrix} \mathbf{P}') \ge \text{rank} \left( \mathbf{A} \begin{pmatrix} \mathbf{B}_1 : \mathbf{B}_2 \end{pmatrix} \mathbf{P}' \mathbf{P} \begin{pmatrix} \mathbf{B}_1^{-1} \\ \mathbf{O} \end{pmatrix} \right) = \text{rank} (\mathbf{A} \mathbf{I}_n) = \text{rank} (\mathbf{A}). $$ Thus $\text{rank}(\mathbf{A}) = \text{rank} (\mathbf{A} \mathbf{B})$. Proof for the other half of the statement follows the same argument.
Example: 2019 qual. exam Q1.
A = [1 -2 -2; 3 -6 -6]
rank(A)
# this B does not have full row rank
B = [2 2; 1 0; 0 1]
rank(A * B)
# B has full row rank
B = [2 2 0; 1 0 0; 0 1 1]
# A * B preserves rank of A
rank(A * B)
Rank-nullity theorem¶
The nullity of a matrix $\mathbf{A}$ is the dimension of its null space $$ \text{nullity}(\mathbf{A}) = \text{dim}(\mathcal{N}(\mathbf{A})). $$
Let $\mathbf{A} \in \mathbb{R}^{m \times n}$, then $$ \text{rank}(\mathbf{A}) + \text{nullity}(\mathbf{A}) = n. $$
Proof (optional): Denote $\nu = \text{nullity}(\mathbf{A}) = \text{dim}(\mathcal{N}(\mathbf{A}))$. Let $\mathbf{X} \in \mathbb{R}^{n \times n}$ be $$ \mathbf{X} = \begin{pmatrix} \mathbf{X}_1 : \mathbf{X}_2 \end{pmatrix}, $$ where columns of $\mathbf{X}_1 \in \mathbb{R}^{n \times \nu}$ form a basis of $\mathcal{N}(\mathbf{A})$ and columns of $\mathbf{X}_2 \in \mathbb{R}^{n \times (n - \nu)}$ extend those in $\mathbf{X}_1$ to be a basis of $\mathbb{R}^n$. We show columns of $\mathbf{A} \mathbf{X}_2$ form a basis of $\mathcal{C}(\mathbf{A})$. Thus $\text{rank}(\mathbf{A}) = \text{dim}(\mathcal{C}(\mathbf{A})) = n - \nu$.
(1) To show that columns of $\mathbf{A} \mathbf{X}_2$ are linearly independent. Assume $\mathbf{A} \mathbf{X}_2 \mathbf{v} = \mathbf{0}$. Then $\mathbf{X}_2 \mathbf{v} \in \mathcal{N}(\mathbf{A}) = \mathcal{C}(\mathbf{X}_1)$. Thus $\mathbf{X}_2 \mathbf{v} = \mathbf{X}_1 \mathbf{u}$ for some $\mathbf{u}$, or equivalently, $$ \begin{pmatrix} \mathbf{X}_1 : \mathbf{X}_2 \end{pmatrix} \begin{pmatrix} -\mathbf{u} \\ \mathbf{v} \end{pmatrix} = \mathbf{0}_n. $$ Since the matrix $\begin{pmatrix} \mathbf{X}_1 : \mathbf{X}_2 \end{pmatrix}$ is non-singular, we must have $\mathbf{u}=\mathbf{0}$ and $\mathbf{v}=\mathbf{0}$. This shows that $\mathbf{v}=\mathbf{0}$ whenever $\mathbf{A} \mathbf{X}_2 \mathbf{v} = \mathbf{0}$. So the columns of $\mathbf{A} \mathbf{X}_2$ are linearly independent.
(2) Next we show the columns of $\mathbf{A} \mathbf{X}_2$ span $\mathcal{C}(\mathbf{A})$ by showing $\mathcal{C}(\mathbf{A} \mathbf{X}_2) \subseteq \mathcal{C}(\mathbf{A})$ and $\mathcal{C}(\mathbf{A} \mathbf{X}_2) \supseteq \mathcal{C}(\mathbf{A})$. One direction $\mathcal{C}(\mathbf{A} \mathbf{X}_2) \subseteq \mathcal{C}(\mathbf{A})$ is easy. To show the other direction $\mathcal{C}(\mathbf{A}) \subseteq \mathcal{C}(\mathbf{A} \mathbf{X}_2)$, let $\mathbf{w} \in \mathcal{C}(\mathbf{A})$. Then $\mathbf{w} = \mathbf{A} \mathbf{y}$ for some vector $\mathbf{y}$. Because $\mathbf{y} \in \mathbb{R}^n$, which is spanned by columns of $\mathbf{X}$, we can write $\mathbf{y} = \mathbf{X}_1 \mathbf{z}_1 + \mathbf{X}_2 \mathbf{z}_2$ for some vectors $\mathbf{z}_1$ and $\mathbf{z}_2$. Thus $\mathbf{w} = \mathbf{A} \mathbf{X}_1 \mathbf{z}_1 + \mathbf{A} \mathbf{X}_2 \mathbf{z}_2 = \mathbf{A} \mathbf{X}_2 \mathbf{z}_2 \in \mathcal{C}(\mathbf{A} \mathbf{X}_2)$. This proves $\mathcal{C}(\mathbf{A}) \subseteq \mathcal{C}(\mathbf{A} \mathbf{X}_2)$.
A = [1 -2 -2; 3 -6 -6]
rank(A)
nullity = size(nullspace(A), 2)
Fundamental theorem of ranks (important)¶
$\text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}') = \text{rank}(\mathbf{A}'\mathbf{A}) = \text{rank}(\mathbf{A}\mathbf{A}')$.
Proof: $\text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}')$ by definition of rank (row rank = column rank = rank). Early on we showed $\mathcal{N}(\mathbf{A}'\mathbf{A}) = \mathcal{N}(\mathbf{A})$. Thus $\text{nullity}(\mathbf{A}'\mathbf{A}) = \text{nullity}(\mathbf{A})$. Then by the rank-nullity theorem, $\text{rank}(\mathbf{A}'\mathbf{A}) = \text{rank}(\mathbf{A})$.
Matrices of form $\mathbf{A}'\mathbf{A}$ or $\mathbf{A}\mathbf{A}'$ are called the Gram matrix or Gramian matrix.
A = [1 -2 -2; 3 -6 -6]
rank(A)
A'A
rank(A'A)
A * A'
rank(A * A')
Rank factorization (important)¶
Let $\mathbf{A} \in \mathbb{R}^{m \times n}$ with rank $r \ge 1$. The product $\mathbf{A} = \mathbf{C} \mathbf{R}$, where $\mathbf{C} \in \mathbb{R}^{m \times r}$ and $\mathbf{R} \in \mathbb{R}^{r \times n}$ is called a rank decomposition or rank factorization of $\mathbf{A}$.
Visualize (TODO): $$ \mathbf{A} = \begin{pmatrix} | & & | \\ \mathbf{c}_1 & \cdots & \mathbf{c}_r \\ | & & | \end{pmatrix} \begin{pmatrix} - & \mathbf{r}_1' & - \\ & \vdots & \\ - & \mathbf{r}_r' & - \end{pmatrix}. $$
Existence of rank factorization. Any non-null matrix has a rank decomposition. To construct one, let columns of \begin{eqnarray*} \mathbf{C} = \begin{pmatrix} \mid & & \mid \\ \mathbf{c}_1 & \cdots & \mathbf{c}_r \\ \mid & & \mid \end{pmatrix} \end{eqnarray*} be a basis of $\mathcal{C}(\mathbf{A})$. Then $\mathcal{C}(\mathbf{A}) \subseteq \mathcal{C}(\mathbf{C})$. Thus there exists $\mathbf{R}$ such that $\mathbf{A} = \mathbf{C} \mathbf{R}$.
Is rank factorization unique? No. $\mathbf{A} = \mathbf{C} \mathbf{R} = (\mathbf{C} \mathbf{M}) (\mathbf{M}^{-1} \mathbf{R})$ for any non-singular matrix $\mathbf{M}^{r \times r}$.
Suppose $\mathbf{A}$ has column rank $r > 0$ and rank factorization $\mathbf{A} = \mathbf{C} \mathbf{R}$. Then \begin{eqnarray*} & & \text{row rank of } \mathbf{A} \\ &=& \text{column rank of } \mathbf{A}' \\ &=& \text{column rank of } \mathbf{R}' \mathbf{C}' \\ &\le& \text{column rank of } \mathbf{R}' \quad \quad (\text{right matrix multiplication shrinks column space}) \\ &\le& r \\ &=& \text{column rank of } \mathbf{A}. \end{eqnarray*} Now applying the above conclusion to matrix $\mathbf{A}'$, we have $$ \text{row rank of } \mathbf{A}' \le \text{column rank of } \mathbf{A}', $$ i.e., $$ \text{column rank of } \mathbf{A} \le \text{row rank of } \mathbf{A}. $$ Thus we have a proof of the famous result $\text{rank}(\mathbf{A}) = \text{rank}(\mathbf{A}')$.
Let $\text{rank}(\mathbf{A}) = r$ and $\mathbf{A} = \mathbf{C} \mathbf{R}$ be a rank factorization. Then
- $\text{rank}(\mathbf{C}) = \text{rank}(\mathbf{R}) = r$,
$\mathcal{C}(\mathbf{A}) = \mathcal{C}(\mathbf{C})$, $\mathcal{C}(\mathbf{A}') = \mathcal{C}(\mathbf{R}')$ and $\mathcal{N}(\mathbf{A}) = \mathcal{N}(\mathbf{R})$.
Proof of 1: $r = \text{rank}(\mathbf{A}) = \text{rank}(\mathbf{C}\mathbf{R}) \le \text{rank}(\mathbf{C}) \le r$. Thus $\text{rank}(\mathbf{C}) = r$. Similarly $\text{rank}(\mathbf{R}) = r$.
Proof of 2 (optional): $\mathcal{C}(\mathbf{A}) \subseteq \mathcal{C}(\mathbf{C})$ is trivial. Suppose $\mathcal{C}(\mathbf{C})$ is strictly larger than $\mathcal{C}(\mathbf{A})$. Then there exists vector $\mathbf{v} \in \mathcal{C}(\mathbf{C})$ that is not a linear combination of columns of $\mathbf{A}$. Let $\mathbf{u}_1, \ldots, \mathbf{u}_r$ be a basis of $\mathcal{C}(\mathbf{A})$. Then the $r+1$ vectors $\mathbf{u}_1, \ldots, \mathbf{u}_r, \mathbf{v}$ is an independent set in $\mathcal{C}(\mathbf{C})$, contadicting the fact $\text{rank}(\mathbf{C}) = r$. Therefore we must have $\mathcal{C}(\mathbf{A}) = \mathcal{C}(\mathbf{C})$. Similar argument shows $\mathcal{C}(\mathbf{A}') = \mathcal{C}(\mathbf{R}')$.
To show $\mathcal{N}(\mathbf{A}) = \mathcal{N}(\mathbf{R})$, one direction $\mathcal{N}(\mathbf{A}) \supseteq \mathcal{N}(\mathbf{R})$ is trivial (why?). To show the other direction, \begin{eqnarray*} & & \mathbf{x} \in \mathcal{N}(\mathbf{A}) \\ &\Rightarrow& \mathbf{A} \mathbf{x} = \mathbf{0} \\ &\Rightarrow& \mathbf{C} \mathbf{R} \mathbf{x} = \mathbf{0} \\ &\Rightarrow& \mathbf{R} \mathbf{x} \in \mathcal{N}(\mathbf{C}). \end{eqnarray*} But by the rank-nullity theorem, $\text{nullity}(\mathbf{C}) = r - \text{rank}(\mathbf{C}) = 0$. Thus $\mathbf{R} \mathbf{x} = \mathbf{0}$, that is $\mathbf{x} \in \mathcal{N}(\mathbf{R})$. We have shown $\mathcal{N}(\mathbf{A}) \subseteq \mathcal{N}(\mathbf{R})$.
A = [1 -2 -2; 3 -6 -6]
C = [1; 3]
R = [1 -2 -2]
C * R