Recall three types of elementary row operations

$R_i \leftrightarrow R_j$
$R_i \rightarrow \lambda R_i$ where $\lambda \neq 0$
$R_i \rightarrow R_i + \lambda R_j$

Now that we’ve studied matrix multiplication we can state the fact that performing an elementary row operation on $A \in M_{2 \times 2}$ can actually be described using matrix multiplication.

Elementary Matrices

Definition

An $m \times n$ elementary matrix obtained from $I_n$ by performing an elementary row operation of type I, II or III.

Example

Applying the three types of elementary row operations results in the following matrices

$$ \begin{align*} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \rightarrow E_1 &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ E_2 &= \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix} \\ E_3 &= \begin{pmatrix} 1 & 0 \\ \lambda & 1 \end{pmatrix} \end{align*} $$

This leads to the next theorem

Theorem

Let $E$ be the elementary matrix obtained from $I_n$ by performing row operation $\mathcal{R}(E = E(\mathcal{R})))$ for any $A \in M_{m \times n}$, the product $$ \begin{align*} E(\mathcal{R}) \cdot A_{m \times n} \end{align*} $$ is equal to the matrix obtained from $A$ by performing $\mathcal{R}$.

Example

Let’s apply the elementary matrices on the following given matrix

$$ \begin{align*} E_1 \begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ E_2 \begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ \lambda c & \lambda d \end{pmatrix} \\ E_3 \begin{pmatrix} a & b \\ c & d \end{pmatrix} &=\begin{pmatrix} 1 & 0 \\ \lambda & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c + \lambda a & d + \lambda b \end{pmatrix} \end{align*} $$

RREF by Matrix Multiplication

Since we can perform elementary row operations by matrix multiplication, then we can possibly see how we can put a matrix in reduced row echelon by multiplication. But first, there is an observation

Corollary

Each elementary matrix is invertible. $$ \begin{align*} &\mathcal{R}: R_i \leftrightarrow R_j \quad \quad \quad \quad \ \ \mathcal{R}^{-1}: R_j \leftrightarrow R_i \\ &\mathcal{R}: R_i \rightarrow \lambda R_i \quad \quad \quad \quad \mathcal{R}^{-1}: R_i \rightarrow \frac{1}{\lambda} R_i \\ &\mathcal{R}: R_i \rightarrow R_i + \lambda R_j \quad \quad \mathcal{R}^{-1}: R_i \rightarrow R_i - \lambda R_j \end{align*} $$

Proof: apply the elementary row operation by multiplying by $E(\mathcal{R})$ and then apply the inverse again by multiplying by $E(\mathcal{R}^{-1})$. The result is the identity matrix. In other words, $E(\mathcal{R})E(\mathcal{R}^{-1}) = I_n$.

Theorem

For every $A \in M_{m \times n}$, there is a finite set of elementary matrices $E_1,...,E_k \in M_{m \times n}$ such that $E_k ... E_2E_1$ is in RREF.

Theorem

$A \in M_{m \times n}$ is invertible if and only iff there is a finite set of elementary matrices $E_1,...,E_k \in M_{n \times n}$ such that $E_k...E_2E_1A = I_n$. $(A \ | \ I_n)$.

Note here that from the expression above we can see that $A^{-1} = E_k...E_1$ and $A = E_1^{-1}E_2^{-1}...E^{-1}_{k-1}E^{-1}_{k}$

Corollary

$A$ is invertible if and only iff it can be written as a product of elementary matrices.

The Rank of a Matrix

Recall that the rank of $T: V \rightarrow W$ is $\dim(R(T))$.

Definition

The rank of $A \in M_{n \times n}$, $rank(A)$ is the rank of $L_A: \mathbf{R}^m \rightarrow \mathbf{R}^n$.

This definition is kind of awkward and instead we want to find an expression for rank($A$) in terms of $A$ itself and not have to rely on $L_A$. To figure this out, we need the following result

Proposition

If $B \in M_{n \times n}$ is invertible, then rank($BA$) = rank($A$).

So multiplication by $B$ doesn’t change the rank.

Proof:
By definition $rank(BA)$ is the rank of the linear map $L_{BA}$. But the rank of a linear map is the dimension of its range and so

$$ \begin{align*} rank(BA) &= rank(L_{BA}) \\ &= \dim(R(L_{BA})). \end{align*} $$

By definition, to get the range, we just apply the linear map on $v \in \mathbf{R}^n$. This range is a subset of $\mathbf{R}^m$. Applying the linear map is just multiplying $v$ by the matrices $A$ and $B$. So we can see this below:

$$ \begin{align*} \dim(R(L_{BA})) &= \{L_{BA}(v) \ | \ v \in \mathbf{R}^n \} \subset \mathbf{R}^m \\ &= \{BA(v) \ | \ v \in \mathbf{R}^n\} \\ &= \{B(A(v)) \ | \ v \in \mathbf{R}^n\} \\ &= \{B(A(v)) \ | \ v \in \mathbf{R}^n\} \end{align*} $$

This is equivalent to applying the linear map $L_{B}$ on the set that we get from applying $A$ as follows

$$ \begin{align*} \dim(R(L_{BA})) &= L_B(\{A(v) \ | \ v \in \mathbf{R}^n\}) \\ \end{align*} $$

But this internal set is the range of the linear map $L_A$, $R(L_A)$. So what we want to show is that applying $L_B$ doesn’t change the dinemsion of $(R(L_A))$ or multiplying by the matrix $B$ above, doesn’t change anything about the dimension of the internal set.

The idea is simple but subtle. We know that $L_B$ is a map from $\mathbf{R}^m$ to $\mathbf{R}^m$. But here, $L_B$ is not acting on $\mathbf{R}^m$ but rather $R(L_A)$ (the internal set)$. The range of$L_A$is a subset of$\mathbf{R}^m$$. So define the map

$$ \begin{align*} L_B&: \mathbf{R}^m \rightarrow \mathbf{R}^m \\ \tilde{L_B}&: R(L_A) \rightarrow L_B(R(L_A)) \end{align*} $$

We claim this new map is invertible. It’s onto because we restricted the target to the image $R(L_A)$ and it’s one-to-one because $B$ is invertible. Therefore, the dimension of the domain and the target are the same (It’s an invertible map!). In other words,

$$ \begin{align*} \dim(R(L_A)) = \dim(L_B(R(L_A))) \end{align*} $$

But we know that $\dim(R(L_A))$ is the rank of $A$. and $\dim(L_B(R(L_A))$ is the rank of $L_{BA}$ so $rank(A) = rank(BA) \ \blacksquare$.

The Rank of a Matrix

So now we can go back to our original goal of finding an expression for finding the rank of a matrix $A$ without going back to the rank of the linear map $L_A$. We just proved that $\text{rank}(BA) = \text{rank}(A)$. for any invertible matrix $B$. We further studied earlier that elementary matrices are invertible. So multiplying $A$ by a set of elementary matrices will not change its rank. So we can get $A$ in RREF without its rank changing. Based on this we have the following corollaries:

Corollary

Elementary row operations don't change rank.

and

Corollary

$$ \begin{align*} rank(A) = rank(RREF(A)) \end{align*} $$

Why do we want RREF? because it’s easy to read off and we can easily figure out the dimension easily from seeing a matrix in its RREF.

Example

What is the range of the following matrix?

$$ \begin{align*} A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \end{align*} $$

We know the rank of $A$. It is the dimension of the range of $L_A$ or $\dim(R(L_A))$. Notice for the above matrix, if we multiply $A$ by a vector $v$, the result is a linear combination of the columns of $A$. Recall

$$ \begin{align*} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = x \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + y \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + z \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + w \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \end{align*} $$

In other words, the columns span the range. Here, we see that we have 3 non-zero columns and they are linearly independent. so $rank(A) = \dim(R(L_A)) = 3$. This works here but it’s not generally true!! we’re missing something … so let’s clarify with another example

Example

What is the range of the following matrix?

$$ \begin{align*} A = \begin{pmatrix} 1 & 0 & 3 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{align*} $$

It’s still true that the range of this is still spanned by the columns (always true). But the rank of $A$ in general is the number of columns with leading entries in $REF$ of $A$.

Nullity and The Dimension Theorem

So we know that $\text{nullity}(A) = \dim(N(A))$. We found a basis for the null space of $A$ by solving $Ax = 0$ and finding all the solutions and then writing a set that spans that solution set. From the basis we knew the dimension of the null space. Specifically when we solved $Ax = 0$, it was the number of columns without leading entries. So we can write $\text{nullity}(A) = \dim(N(A)) =$ # of columns without leading entries.

So now if we put together the number of columns without leading entries (nullity of $A$) and the number of columns with leading entries (rank of $A$), then we get $n$. This is basically the dimension theorem.

References

Math416 by Ely Kerman