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Dimension of a Vector Space1

Dated: 03-06-2025

The dimensions of a vector space1 is the number of vectors2 in the basis \(B\) of that given vector space1 and is denoted as \(\dim \mathbb V\).

Theorem

If a vector space1 \(\mathbb V\) has a basis \(B = \{\vec b_1, \ldots, \vec b_n\}\), then any set3 in \(\mathbb V\) containing more than \(n\) vectors2 must be linearly dependent.4

Finite and Infinite Dimensional Vector Spaces

If a vector space1 \(\mathbb V\) is spanned by a finite basis then \(\mathbb V\) is said to be finite dimensional and otherwise if we unable to find a finite basis then \(\mathbb V\) is said to be infinite dimensional.

Note

  • The dimension of a null vector space1 is defined to be \(0\).
  • Every finite dimensional vector space1 contains a basis.

Dimensions of some vector spaces1

\(\dim (\mathbb R^n) = n\)
\(\dim (\mathbb P_n) = n + 1\)
\(\dim (M_{m \times n}) = mn\)

Example

Let \(\mathbb W\) be the subspace1 of the set3 of all matrices5 defined by

\[ \mathbb W = \left\{ \mathbf A= \begin{bmatrix} a & b\\ c & d \end{bmatrix} : 2a - b + 3c + d = 0 \right\} \]
\[\because d = -2a + b - 3c\]
\[ \mathbf A= \begin{bmatrix} a & b\\ c & d \end{bmatrix} = \begin{bmatrix} a & b\\ c & -2a + b - 3c \end{bmatrix} \]
\[ \mathbf A = \begin{bmatrix} a & 0 \\ 0 & -2a \end{bmatrix} + \begin{bmatrix} 0 & b \\ 0 & b \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ c & -3c \end{bmatrix} \]
\[ = a \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 1 & -3 \end{bmatrix} \]
\[ \mathbf A_1 = \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix}, \, \mathbf A_2 = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}, \text{and } \mathbf A_3 = \begin{bmatrix} 0 & 0 \\ 1 & -3 \end{bmatrix} \]
\[= a \mathbf A_1 + b \mathbf A_2 + c \mathbf A_3\]

Therefore, \(\{\mathbf A_1, \mathbf A_2, \mathbf A_3\}\) is the spanning set3 for \(\mathbb W\).
Now need to check if it is basis or not.

\[a \mathbf A_1 + b \mathbf A_2 + c \mathbf A_3 = 0 \implies a = b = c = 0\]
\[ a \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 1 & -3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]
\[ \begin{bmatrix} a & 0 \\ 0 & -2a \end{bmatrix} + \begin{bmatrix} 0 & b \\ 0 & b \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ c & -3c \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]
\[ \begin{bmatrix} a & b \\ c & -2a+b-3c \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]
\[\implies a = b = c =0\]
\[\therefore \dim(\mathbb W) = 3 \quad \because n\{\mathbf A_1, \mathbf A_2, \mathbf A_3\} = 3\]

Theorem

The pivot columns6 of a matrix5 \(\mathbf A\) form a basis for \(\text{Col } \mathbf A\).

Procedure

Imagine a set1 of vectors1 \(S = \{\vec v_1, \vec v_2, \ldots, \vec v_k\} \in \mathbb R^n\). The following procedure produces \(b \subset S\) which is basis for \(span(S)\) and expresses \(\vec v \in S - b\) as linear combinations7 of \(\vec p \in b\).

Step 1

Form a matrix5 \(A\) with \(\vec v_1, \vec v_2, \ldots, \vec v_k\) as its column vectors.

Step 2

Reduce \(A\) to reduced echelon form8 \(R\) and let \(w_1, w_2, \ldots, w_k\) be the column vectors of \(R\).

Step 3

Identify the column vectors in \(R\) with leading \(1\)'s. The corresponding column vectors in \(A\) are the basis for \(span(S)\).

Step 4

Express each column vector of \(R\) which does not contain the leading entries, as linear combinations7 of the previous column vectors which do contain the leading entries.

Example

Find the subset3 of following vectors1 that form the basis for the vector space1 spanned by them.

\[\vec v_1 = \langle1, -2, 0, 3\rangle\]
\[\vec v_2 = \langle2, -5, -3, 6\rangle\]
\[\vec v_3 = \langle0, 1, 3, 0\rangle\]
\[\vec v_4 = \langle2, -1, 4, -7\rangle\]
\[\vec v_5 = \langle5, -8, 1, -2\rangle\]

\(A\) matrix:5

\[ \begin{array}{ccccc} \vec v_1 & \vec v_2 & \vec v_3 & \vec v_4 & \vec v_5\\ \Big\downarrow & \Big\downarrow & \Big\downarrow & \Big\downarrow & \Big\downarrow\\ \end{array} \]
\[ \begin{bmatrix} 1 & 2 & 0 & 2 & 5 \\ -2 & -5 & 1 & -1 & -8 \\ 0 & -3 & 3 & 4 & 1 \\ 3 & 6 & 0 & -7 & 2 \\ \end{bmatrix} \]

\(B\) matrix:5

\[ \begin{array}{ccccc} \vec w_1 & \vec w_2 & \vec w_3 & \vec w_4 & \vec w_5\\ \Big\downarrow & \Big\downarrow & \Big\downarrow & \Big\downarrow & \Big\downarrow\\ \end{array} \]
\[ \left[ \begin{array}{ccccc} 1 & 0 & 2 & 0 & 1 \\ 0 & 1 & -1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right] \]

The leading entries occur in columns 1, 2 and 4 so \(\{\vec w_1, \vec w_2, \vec w_4\}\) is a basis for the column space of \(B\) and consequently \(\{\vec v_1, \vec v_2, \vec v_4\}\) is the basis for column space of \(A\).

\[\vec w_3 = 2 \vec w_1 - \vec w_2\]
\[\vec w_5 = \vec w_1 + \vec w_2 + \vec w_4\]
\[\vec v_3 = 2 \vec v_1 - \vec v_2 \quad \text{and} \quad \vec v_5 = \vec v_1 + \vec v_2 + \vec v_4\]

Subspaces of a Finite-dimensional Space

Theorem

Let \(\mathbb H\) be a subspace1 of a finite dimensional vector space \(\mathbb V\). Any linearly independent4 set3 in \(\mathbb H\) can be expanded to a basis for \(\mathbb H\) and \(\dim(\mathbb H) \le \dim(\mathbb V)\).

Note

The dimension of \(\text{Nul } A\) is the number of free variables in the equation \(A \vec x = \vec 0\) and the dimension of \(\text{Col } A\) is the number of pivot columns6 in \(A\).

References

Read more about notations and symbols.

  1. Read more about vector spaces. ↩↩↩↩↩↩↩↩↩↩↩↩↩↩

  2. Read more about vectors. ↩↩

  3. Read more about sets. ↩↩↩↩↩

  4. Read more about linear dependency. ↩↩

  5. Read more about matrices. ↩↩↩↩↩

  6. Read more about pivot columns and positions. ↩↩

  7. Read more about linear combinations. ↩↩

  8. Read more about reduced echelon forms. ↩