Linear Independence: Examples

Example

Determine if the vectors

\[\Vect{a}_1=(1,2,5)\quad \text{and} \quad \Vect{a}_2=(2,-1,4)\]

of \(\RNrSpc{3}\) are linearly independent.

Solution

We need to examine the possible solutions of the vector equation

\(t_1 \Vect{a}_1 + t_2 \Vect{a}_2\)\(=\)\(\Vect{0}\)
\[ t_1 \left[ \begin{array}{c} 1 \\ 2 \\ 5 \end{array} \right]\ +\ t_2 \left[ \begin{array}{r} 2 \\ -1 \\ 4 \end{array} \right]\ =\ \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] \]

This homogeneous system of linear equations has as its reduced row echelon form

\[ \left[ \begin{array}{cc|c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] \]

This says that \(t_1=t_2=0\) is the only solution of the given vector equation. We conclude that the set of vectors \(\Set{ \Vect{a}_1 , \Vect{a}_2 }\) is linearly independent.

Example

Determine if the vectors

\[\Vect{a}_1=(1,2,5), \quad \Vect{a}_2=(2,-1,4),\quad \text{and}\quad \Vect{a}_3=(0,5,6)\]

of \(\RNrSpc{3}\) are linearly independent. If they are not, express one vector as a linear combination of the others.

Solution

We need to examine the possible solutions of the vector equation

\(t_1 \Vect{a}_1 + t_2 \Vect{a}_2 + t_3 \Vect{a}_3\)\(=\)\(\Vect{0}\)
\[ t_1 \left[ \begin{array}{c} 1 \\ 2 \\ 5 \end{array} \right]\ +\ t_2 \left[ \begin{array}{r} 2 \\ -1 \\ 4 \end{array} \right]\ +\ t_3 \left[ \begin{array}{r} 0 \\ 5 \\ 6 \end{array} \right] =\ \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] \]

This homogeneous system of linear equations has as its reduced row echelon form

\[ \begin{array}{rrr|r} 1 & 0 & 2 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \]

Now we see that this system of linear equation has infinitely many solutions, as \(t_3\) can be chosen arbitrarily. We conclude that the vectors \(\Set{ \Vect{a}_1 , \Vect{a}_2 , \Vect{a}_3 }\) fail to be linearly independent: they are linearly dependent.

To express one vector as a linear combination of the others, we may use any of the solutions of the above system of linear equations. For example, \(t_3 = 4\) yields

\(t_1\)\(=\)\(-2\cdot 4 = -8\)
\(t_2\)\(=\)\(4\)

This means that

\(-8\cdot \Vect{a}_1 + 4\cdot \Vect{a}_2 + 4\cdot \Vect{a}_3\)\(=\)\(\Vect{0}\)
\(\Vect{a}_2\)\(=\)\(2\cdot \Vect{a}_1 - \Vect{a}_3\)

Notice that we could just as well have expressed \(\Vect{a}_1\) as a linear combination of \(\Vect{a}_2\) and \(\Vect{a}_3\):

\(\Vect{a}_1\)\(=\)\(\dfrac{1}{2}\cdot (\Vect{a}_2 +\Vect{a}_3)\)

Similarly, \(\Vect{a}_3\) is a linear combination of \(\Vect{a}_1\) and \(\Vect{a}_2\).