# Dimension Theorem and Rank Theorem

## Definition

Let $A\in \mathfrak{M}_{m\times n}(\mathbb{R})$ be a matrix. We define the column rank of $A$ as dimension of $\langle [A]^1, \ldots, [A]^n \rangle$, i.e., span of column vectors, where $[A]^j\in \mathbb{R}^m$ denotes $j$-th column of the matrix $A$. Similarly, we define the row rank of $A$ as dimension of $\langle [A]_1, \ldots, [A]_m\rangle$, where $[A]_i\in\mathbb{R}^n$ denotes a $i$-th row of $A$.

## Remark

Note that column space $\langle [A]^1, \ldots, [A]^n\rangle$ and row space $\langle [A]_1, \ldots , [A]_m\rangle$ is subspace of $\mathbb{R}^m$ and $\mathbb{R}^n$, respectively. Moreover column space is equal to the image of $L_A$, where $L_A: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is a linear map such that $\mathbf{x}\mapsto A\mathbf{x}$.

\begin{align*} \text{im} L_A &= \{A\mathbf{x} \mid \mathbf{x}\in\mathbb{R}^n \} \\ &=\{x_1[A]^1 + \cdots +x_n[A]^n \mid (x_1, \ldots, x_n)\in \mathbb{R}^n \} \\ &=\langle [A]^1, \ldots, [A]^n \rangle \end{align*}

## Dimension Theorem

Let $(V,+,\cdot)$, $(W,+,\cdot)$ be finite dimensional vector spaces of $F$. Let $\varphi: V\rightarrow W$ be a linear map. Then $\dim V= \dim \ker \varphi + \dim \text{im}\varphi$.

<proof>

Suppose that $\dim V = n$. Let $S=\{\mathbf{v}_1, \ldots, \mathbf{v}_k \}$ be a basis for $\ker\varphi$. By the basis extension theorem, we get a basis $\{ \mathbf{v}_1, \ldots, \mathbf{v}_k, \mathbf{v}_{k+1}, \ldots, \mathbf{v}_n\}$ for $V$ by extending the set S. It suffices to show that $\{ \varphi(\mathbf{v}_{k+1}), \ldots, \varphi(\mathbf{v}_n)\}$ is a basis for im$\varphi$.

Suppose that

$$$a_{k+1}\varphi(\mathbf{v}_{k+1}) + \cdots + a_{n}\varphi(\mathbf{v}_n) =\mathbf{0}.$$$

By linearity of $\varphi$,

$$$a_{k+1}\varphi(\mathbf{v}_{k+1}) + \cdots + a_{n}\varphi(\mathbf{v}_n)=\varphi\left(\sum_{i=k+1}^na_i\mathbf{v}_i\right) = \mathbf{0}.$$$

That is $\sum_{i=k+1}^na_i\mathbf{v}_i\in \ker\varphi$. Since $S$ is a basis for $\ker\varphi$,

$$$\sum_{i=k+1}^na_i\mathbf{v}_i = \sum_{j=1}^k b_j\mathbf{v}_j \iff \sum_{i=k+1}^na_i\mathbf{v}_i - \sum_{j=1}^k b_j\mathbf{v}_j = \mathbf{0}.$$$

Since $\{\mathbf{v}_1, \ldots, \mathbf{v}_n\}$ is a basis for $V$, $a_i=0, b_j=0$ for all $i=1,\ldots, k$ and $j=k+1, \ldots, n$. Thus, $\{ \varphi(\mathbf{v}_{k+1}), \ldots, \varphi(\mathbf{v}_n)\}$ is linearly independent subset.

Let $\mathbf{w}\in \text{im}\varphi$ be given. Then $\mathbf{w} = \varphi(\mathbf{v})$ for some $\mathbf{v}\in V$. Since $\{\mathbf{v}_1, \ldots, \mathbf{v}_n\}$ is a basis for $V$, we get a unique linear combination of $\mathbf{v}=\sum_{i=1}^n a_i \mathbf{v}_i$.

\begin{align*} \varphi(\mathbf{v})&=\varphi\left(\sum_{i=1}^n a_i \mathbf{v}_i\right) \\ &=\sum_{i=1}^na_i\varphi(\mathbf{v}_i) \\ &=\sum_{i=k+1}^n a_i\varphi(\mathbf{v}_i) \\ &\in \text{span}\{ \varphi(\mathbf{v}_{k+1}, \ldots, \varphi(\mathbf{v}_n)\} \end{align*}

The last equality holds since $\mathbf{v}_j \in \ker\varphi$, i.e., $\varphi(\mathbf{v}_j)=\mathbf{0}$ for all $j=1,\ldots,k$. Thus, $\{\varphi(\mathbf{v}_{k+1}), \ldots, \varphi(\mathbf{v}_n)\}$ generates im$\varphi$.

$\tag*{\square}$

## Rank Theorem

For any $m\times n$ matrix $A$, the column rank of $A$ is equal to the row rank of $A$.

<proof> Since

\begin{align} \begin{split} n&=\dim \ker L_A + \dim \text{im}L_A \\ &=\dim \ker L_A + (\text{column rank of } A) \end{split} \label{eq:1} \end{align}

by dimension theorem, it suffices to show that

$$$n= \dim \ker L_A + (\text{row rank of } A) \label{eq:2}$$$

Let $R$ be the reduced row-echelon form of $A$. We know that $R$ is unique. In general, the reduced row-echelon form is

$R= \begin{pmatrix} 0 &1 & * & 0 & * & 0 & * & \cdots \\ 0&0&0&1&*&0&*&\cdots \\ 0&0&0&0&*&1&*&\cdots \\ 0&0&0&0&0&0&0&\cdots\\ \vdots&\vdots&\vdots&\vdots &\vdots &\vdots&\vdots&\ddots \\ 0&0&0&0&0&0&0&\cdots \end{pmatrix}$

We know that elementary row operation does not change the row space of $A$, thus row space of $R$ is equal to the row space of $A$. We observe that dimension of the row space is the number of non-zero rows, i.e., the rows with leading one. For the dimension of $\ker L_R$, it is same as the dimension of the solution space of $R\mathbf{x}=\mathbf{0}$. Then its dimension is the number of free variables, i.e., the number of columns with no leading one. Since elementary row operation does not change the solution space of $A\mathbf{x}=0$, we get $\dim \ker L_A = \dim \ker L_R$. Thus, row rank of $A$ is equal to the column rank of $A$ by Equation $\ref{eq:1}$ and $\ref{eq:2}$.

$\tag*{\square}$

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