Question

Could a set of three vectors in \({\mathbb{R}^4}\) span all of \({\mathbb{R}^4}\)? Explain.

What about \(n\) vectors in \({\mathbb{R}^m}\) when \(n\) is less than \(m\)?

Short answer

A set of three vectors in \({\mathbb{R}^4}\) cannot span in \({\mathbb{R}^4}\) because the matrix does not have enough columns to fill all the rows. The set of \(n\) vectors in \({\mathbb{R}^m}\) cannot span \({\mathbb{R}^m}\) for \(m > n\).

Step-by-step solution

Writing the definition of \(A{\bf{x}}\)

The column of matrix \(A\) is represented as \(\left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right]\), and vector x is represented as \(\left[ {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right]\).

According to the definition, the weights in a linear combination of matrix A columns are represented by the entries in vector x.

The matrix equation as a vector equation can be written as shown below:

\(\begin{array}{c}A{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right]\\b = {x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n}\end{array}\)

The number of columns in matrix \(A\) should be equal to the number of entries in vector x so that \(A{\bf{x}}\) can be defined.

Writing the condition for a consistent solution

Consider an \(m \times n\) ordered matrix A. Let \(m = n\), which means that the number of rows is equal to the number of columns.

For \(m = n\), the matrix has a maximum of \(n\) pivot positions that can be filled by \(m\) rows. So, the equation \(A{\bf{x}} = {\bf{b}}\) is consistent in this case.

Checking if the set of three vectors can span \({\mathbb{R}^m}\)

The general vector is defined as \(A = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right)\).

A set of three vectors can be represented as \(A = \left( {\begin{array}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}&{{{\bf{a}}_3}}\end{array}} \right)\).

These vectors are in \({\mathbb{R}^4}\) span. So, there will be four rows.

To have a pivot position in each row, so there should be at least four columns. But in this case, there are only three columns.

Therefore, a set of three vectors in \({\mathbb{R}^4}\) cannot span \({\mathbb{R}^4}\).

Checking if the set of n vectors can span \({\mathbb{R}^m}\)

Consider an \(m \times n\) ordered matrix A. Given that \(m > n\), which means that the number of rows is greater than the number of columns.

For \(m > n\), the matrix has a maximum of \(n\) pivot positions that cannot be filled by \(m\) rows. It means the vectors in \({\mathbb{R}^m}\) cannot span \({\mathbb{R}^m}\).

Therefore, the set of \(n\) vectors in \({\mathbb{R}^m}\) cannot span \({\mathbb{R}^m}\) for \(m > n\).