📂Matrix Algebra

Basis of Row Space, Column Space, and Null Space

Overview¹

The concepts such as row space, column space, null space were created to solve linear systems $A \mathbf{x} = \mathbf{b}$. Linear systems can be solved through basic row operations, and indeed, the row space and null space are invariant under basic row operations, indicating their relationship with linear systems. It is important to note here that the column space is not invariant under basic row operations.

Theorem 1

(a1) Two matrices that are row equivalent have the same row space. In other words, basic row operations do not change the row space.

(b1) Two matrices that are row equivalent have the same null space. In other words, basic row operations do not change the null space.

Proof

(a1)

Let’s assume that two matrices $A,B$ are row equivalent. This means each matrix can be obtained from the other through basic row operations. This implies that each row of the matrices can be obtained as an arbitrary linear combination of the rows of the other matrix. Therefore, by the definition of span, the row spaces of the two matrices are the same.

■

(b1)

Let’s assume that two matrices $A,B$ are row equivalent. Since basic row operations do not change the solutions of linear systems, the solutions of the two linear systems $A\mathbf{x} = \mathbf{0}$ and $B \mathbf{x} = \mathbf{0}$ are the same. Therefore, the null spaces of the two matrices are the same.

■

Example on Column Space

Let’s assume two matrices $A$ and $B$ are as follows.

$$A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix} ,\quad B = \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}$$

Then, by adding $-2$ times the first row to the second row of $A$, we get $B$, which means $A$, and $B$ are row equivalent. However, the column space of $A$ is spanned by $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$, but the column space of $B$ is spanned by $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, so the column spaces of the two matrices are different. It is important to note here that, as can be seen from Theorem 3, while the spaces change, the dimensions do not.

Theorem 2

Let’s call the matrix $R$ a row echelon form. Then, the row vectors with leading 1’s constitute the basis of the row space of $R$, and the column vectors with leading 1’s constitute the basis of the column space of $R$.

Theorem 3

If two matrices $A, B$ are row equivalent, the following holds:

(a3) A necessary and sufficient condition for the column vectors of $A$ to be linearly independent is that the corresponding column vectors of $B$ are linearly independent.

(b3) A necessary and sufficient condition for the column vectors of $A$ to form the basis of the column space of $A$ is that the corresponding column vectors of $B$ are linearly independent.

---

From the above theorems, it can be understood that basic row operations do not change the dimension of the column space, but, as shown in the example above, it’s important to note again that this does not mean the column space itself is unaltered.