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25. Rank

Dated: 05-06-2025

The rank is the maximum number of linearly independent1 columns in a matrix2 \(A\) (or \(A^T\)).

The Row Space

Each row in \(A_{m \times n}\) has \(n\) entries and therefore, can be identified as a vector3 in \(\mathbb R^n\).
The set4 of all linear combinations of these row vectors3 is called the row space of \(A\) and is denoted by \(\text{Row } A\).

Theorem

\[A \sim B \implies \text{Row }A = \text{Row }B\]

The non zero rows of echelon form5 of \(B\) are basis for \(\text{Row } A\) and \(\text{Row B}\).

Theorem

If \(A \sim B\) then

  • \(\{\vec C_1, \vec C_2, \ldots \vec C_n\}\) of \(A\) forms basis for \(\text{Col }A\) \(\iff\) \(\{\vec C_1, \vec C_2, \ldots \vec C_n\}\) of \(B\) forms basis for \(\text{Col }B\).

Example

\[ R = \begin{bmatrix} 1 & -2 & 5 & 0 & 3 \\ 0 & 1 & 3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]

The matrix2 \(R\) is in echelon form.5
The following vectors3 for the basis for \(\text{Row }R\)

\[ \vec r_1 = \begin{bmatrix} 1 & -2 & 5 & 0 & 3 \end{bmatrix} \]
\[ \vec r_2 = \begin{bmatrix} 0 & 1 & 3 & 0 & 0 \end{bmatrix} \]
\[ \vec r_3 = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 \end{bmatrix} \]

And the following vectors3 for the basis for \(\text{Col }R\)

\[ c_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, c_2 = \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, c_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \]

Definition

\(\dim(\text{Col }A) = \dim(\text{Row }A^T)\) is called the rank.
The dimension6 of the null space7 is called the nullity of \(A\).

Rank Theorem

The rank of \(A\) is equal to the pivot positions8 in \(A\).

\[\text{rank }A + \dim(\text{Nul } A) = n\]

Example

\(A_{7 \times 9}\) has a 2 dimensional null space,7 what is the rank of \(A\).

Solution

Since \(A\) has \(9\) columns,

\[\text{rank } A + 2 = 9\]
\[\text{rank } A = 9 - 2 = 7\]

Example

Can \(B_{6 \times 9}\) has a 2 dimensional null space?7

Solution

No, \(B_{6 \times 9}\) has \(9\) columns, each with \(6\) elements. Therefore, the \(\text{Col }B\) exists in \(\mathbb R^6\) and therefore, the dimensions of \(\text{Col } B\) cannot exceed \(6\).

Theorem

  • \(\text{rank }(A_{m \times n})\) is the number of leading variables in the solution \(A \vec x = \vec 0\)
  • \(\text{nullity }(A_{m \times n})\) is the number of parameters in the general solution of \(A \vec x = \vec 0\)

\(\(\text{rank }(A) = \text{rank }(A^T)\)\)

Four Fundamental Matrix Spaces

Fundamental Spaces Dimension
\(\text{Row } A_{m \times n}\) \(r\)
\(\text{Col } A_{m \times n}\) \(r\)
\(\text{Nul } A_{m \times n}\) \(n - r\)
\(\text{Nul } A^T_{m \times n}\) \(m - r\)

Here \(r\) is \(\text{rank }(A)\).

Theorem

The following statements are each equivalent to the statement that \(A_{n \times n}\) is an invertible matrix.9

  • The column vectors3 of \(A\) form basis of \(\mathbb R^n\)
  • \(\text{Col }(A) = \mathbb R^n\)
  • \(\dim(\text{Col }(A)) = n\)
  • \(\text{rank }(A) = n\)
  • \(\text{Nul }(A) = \{\vec 0\}\)
  • \(\dim(\text{Nul }(A)) = n\)

References

Read more about notations and symbols.

  1. Read more about linear independence. ↩

  2. Read more about matrices. ↩↩

  3. Read more about vectors. ↩↩↩↩↩

  4. Read more about sets. ↩

  5. Read more about echelon form. ↩↩

  6. Read more about dimensions. ↩

  7. Read more about null space. ↩↩↩

  8. Read more about pivot positions. ↩

  9. Read more about invertible matrices. ↩