Linear Combinations

Section 5.2 Linear Combinations

Subsection 5.2.1 Writing Linear Combinations

In Section 3.5, I asked you to write some vectors as linear combinations of other vectors. In those calculations, you had to directly solve a system to find the correct coefficients for the linear combinations. In the present section, I want to return to that problem now that I have defined matrices and row reduction as tools. However, I also want to re-interpret the problem.

Subsection 5.2.2 Bases for Spans

In Subsection 3.1.3, I defined bases for linear subspaces; to recap, a basis is a linearly independent set of vectors which spans the subspace. Since the basis spans the subspace, any vector in the subspace can be written as a linear combination of the vectors in the basis. A basis gives a way of describing a subspace.

If \(\{v_1, v_2, v_3\}\) are a basis for a three-dimensional subspace in \(\RR^n\text{,}\) then the vectors in this subspace are precisely all the vectors of the form \(av_1 + bv_2 + cv_3\) where \(a\text{,}\) \(b\) and \(c\) are any real constants. To talk about a specific vector in a subspace, I need to specify the three constants. If I do, I can say that the vector is presenting in terms of the basis.

Let me be more concrete and consider the vector \(v = \begin{pmatrix} 4 \\ 5 \end{pmatrix}\) in \(\RR^2\text{.}\) I can express this as a linear combination of the standard basis.

\begin{equation*} v = 4 e_1 + 5 e_2 \end{equation*}

However, I can express \(v\) in any of the infinitely-many other bases of \(\RR^2\) as well. Here is one example.

\begin{equation*} v = 1 \begin{pmatrix} 4 \\ 0 \end{pmatrix} - 1 \begin{pmatrix} 0 \\ -5 \end{pmatrix} \end{equation*}

Here is yet another example.

\begin{equation*} v = \frac{9}{2} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} -1 \\ 1 \end{pmatrix} \end{equation*}

The problem of finding an expression of a vector as a linear combination of other vectors is the problem of expressing a vector in terms of a given basis. Depending on the situation, different bases for the same subspaces are often useful, and I often want to change the presentation of vectors from one basis to another.

Example 5.2.1.

The subspace \(\Span \left\{ \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 0 \\ -2 \\ 3\end{pmatrix} \right\}\) is a plane in \(\RR^3\text{.}\) The vector \(\begin{pmatrix} -4 \\ -4 \\ 7\end{pmatrix} \) is a vector on this plane. Write this vector in terms of the given basis for the span.

To do this, I need to find the numbers \(a\) and \(b\) that satisfy the following:

\begin{equation*} \begin{pmatrix} -4 \\ -4 \\ 7 \end{pmatrix} = a \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + b \begin{pmatrix} 0 \\ -2 \\ 3 \end{pmatrix}\text{.} \end{equation*}

I’ll write each component of this vector equation separately.

\begin{align*} -4 \amp = -2a + 0b\\ -4 \amp = a + (-2)b\\ 7 \amp = -1a + 3b \end{align*}

This is just a new linear system, with three equations and two variables, \(a\) and \(b\text{.}\) I can solve it using matrices and row reduction. The matrix representing this system is

\begin{equation*} \left( \begin{array}{cc|c} -2 \amp 0 \amp -4 \\ 1 \amp -2 \amp -4 \\ -1 \amp 3 \amp 7 \end{array} \right) \text{.} \end{equation*}

This row reduces to

\begin{equation*} \left( \begin{array}{cc|c} 1 \amp 0 \amp 2 \\ 0 \amp 1 \amp 3 \\ 0 \amp 0 \amp 0 \end{array} \right) \text{.} \end{equation*}

This is solvable and the solution is \(a = 2\) and \(b=3\text{.}\) This lets me write the vector in terms of the basis of the plane.

\begin{equation*} \begin{pmatrix} -4 \\ -4 \\ 7 \end{pmatrix} = 2 \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + 3 \begin{pmatrix} 0 \\ -2 \\ 3 \end{pmatrix} \end{equation*}

If the system wasn’t solvable (if the last row of the matrix turned out to be equivalent to \(0=1\)), then the vector I started with wouldn’t have been on the plane. Only vectors on the plane can be expressed in terms of the basis of the plane.

Example 5.2.2.

Take the basis \(\left\{ \begin{pmatrix} 4 \\ -1 \\ -1 \end{pmatrix}, \begin{pmatrix} -2 \\ 0 \\ -3 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \\ 7 \end{pmatrix} \right\}\) for \(\RR^3\text{.}\) Express the vector \(\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\) in terms of this basis.

To do this, I need to find constants \(a\text{,}\) \(b\) and \(c\) that solve

\begin{equation*} \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = a \begin{pmatrix} 4 \\ -1 \\ -1 \end{pmatrix} + b \begin{pmatrix} -2 \\ 0 \\ -3 \end{pmatrix} + c \begin{pmatrix} 0 \\ 2 \\ 7 \end{pmatrix}\text{.} \end{equation*}

Written in components, this gives the following linear system.

\begin{align*} 3a - 2b + 0c \amp = 1\\ -a + 0b + 2c \amp = 1\\ -a + 3b + 7c \amp = 2 \end{align*}

I translate this into a matrix.

\begin{equation*} \left( \begin{array}{ccc|c} 3 \amp -2 \amp 0 \amp 1 \\ -1 \amp 0 \amp 2 \amp 1 \\ -1 \amp 3 \amp 7 \amp 2 \end{array} \right) \end{equation*}

I row reduce this matrix to get the reduced row-echelon form.

\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp \frac{-1}{2} \\ 0 \amp 0 \amp 1 \amp \frac{1}{2} \end{array} \right) \end{equation*}

I conclude that \(a=0\text{,}\) \(b=\frac{-1}{2}\) and \(c = \frac{1}{2}\text{,}\) which lets me write the vector in terms of the basis.

\begin{equation*} \begin{pmatrix} 1 \\ 1 \\ 2\end{pmatrix} = 0 \begin{pmatrix} 3 \\ -1 \\ -1 \end{pmatrix} - \frac{1}{2} \begin{pmatrix} -2 \\ 0 \\ -3 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 \\ 2 \\ 7 \end{pmatrix} \end{equation*}

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