## 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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