Linear Algebra and Its Applications, Exercise 2.3.7

Exercise 2.3.7. For each of the following, state whether the vector b is in the subspace spanned by w_1, \dotsc, w_l. (Construct a matrix A with w_1, \dotsc, w_l as the columns, and try to solve Ax = b.)

a) w_1 = (1, 1, 0), w_2 = (2, 2, 1), w_3 = (0, 0, 2), b = (3, 4, 5)

b) w_1 = (1, 2, 0), w_2 = (2, 5, 0), w_3 = (0, 0, 2), w_4 = (0, 0, 0), any b

Answer: a) We construct the matrix

A = \begin{bmatrix} 1&2&0 \\ 1&2&0 \\ 0&1&2 \end{bmatrix}

and do Gaussian elimination on the left and right hand sides of Ax = b:

\begin{bmatrix} 1&2&0&\vline&3 \\ 1&2&0&\vline&4 \\ 0&1&2&\vline&5 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&0&\vline&3 \\ 0&0&0&\vline&1 \\ 0&1&2&\vline&5 \end{bmatrix}

\Rightarrow \begin{bmatrix} 1&2&0&\vline&3 \\ 0&1&2&\vline&5 \\ 0&0&0&\vline&1 \end{bmatrix}

From the third equation we have the contradiction 0 = 1 and we conclude that Ax = b has no solution in this case. The vector b = (3, 4, 5) is therefore not in the subspace spanned by w_1, w_2,  and w_3.

b) We construct the matrix

A = \begin{bmatrix} 1&2&0&0 \\ 2&5&0&0 \\ 0&0&2&0 \end{bmatrix}

and do Gaussian elimination on the left and right hand sides of Ax = b:

\begin{bmatrix} 1&2&0&0&\vline&b_1 \\ 2&5&0&0&\vline&b_2 \\ 0&0&2&0&\vline&b_3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&0&0&\vline&b_1 \\ 0&1&0&0&\vline&b_2 - 2b_1 \\ 0&0&2&0&\vline&b_3 \end{bmatrix}

The pivots are in columns 1 through 3 and the basic variables are x_1, x_2, and x_3 with x_4 being a free variable. The first three columns of the original matrix are thus linearly independent; the fourth column is a linear combination of the others. Since the first three columns w_1, w_2, and w_3 are vectors in \mathbf{R}^3 and are linearly independent, they span all of \mathbf{R}^3 and thus any vector b = (b_1, b_2, b_3) can be represented as a linear combination of w_1, w_2, and w_3 using suitable weights.

Going further, from the third equation we have 2x_3 = b_3 or x_3 = b_3/2. From the second equation we have x_2 = b_2 - 2b_1. From the first equation we have

x_1 + 2x_2 = b_1 \rightarrow x_1 + 2(b_2 - 2b_1) = b_1

\rightarrow x_1 + 2b_2 -4b_1 = b_1 \rightarrow x_1 = 5b_1 - 2b_2

The equation Ax = b is equivalent to

\begin{bmatrix} 1&2&0&0 \\ 2&5&0&0 \\ 0&0&2&0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}

\rightarrow x_1 \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ 5 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} + x_4 \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}

Dropping the zero vector w_4 we can therefore represent any vector b in \mathbf{R}^n as a linear combination of w_1, w_2, and w_3 as follows:

\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} = (5b_1 - 2b_2) \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + (b_2 - 2b_1) \begin{bmatrix} 2 \\ 5 \\ 0 \end{bmatrix} + (b_3/2) \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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