## Linear Algebra and Its Applications, Exercise 2.3.14

Exercise 2.3.14. Suppose we have the following matrix

$A = \begin{bmatrix} 1&2&1 \\ 0&0&4 \end{bmatrix}$

How can you extend the rows of $A$ to create a basis for $\mathbf{R}^3$? How can you reduce the columns of $A$ to create a basis for $\mathbf{R}^2$?

Answer: As defined $A$ is in echelon form but has only two pivots, in the first and third columns. We can add another row to $A$ to provide a pivot for the second column:

$A' = \begin{bmatrix} 1&2&1 \\ 0&1&0 \\ 0&0&4 \end{bmatrix}$

Since $A'$ is in echelon form and has pivots in every column, the columns are linearly independent. Since $A'$ also has pivots in every row the rows are also linearly independent. (See exercise 2.3.5.) The three rows also span $\mathbf{R}^3$ with any vector $v = (v_1, v_2, v_3)$ in $\mathbf{R}^3$ expressible as

$v = v_1 \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} + (v_2 - 2v_1) \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + \frac{1}{4} (v_3 - v_1) \begin{bmatrix} 0 \\ 0 \\ 4 \end{bmatrix}$

Since the three rows of $A'$ are linearly independent and span $\mathbf{R}^3$ they form a basis for $\mathbf{R}^3$.

Turning to the columns of $A$, since $A$ has three columns but only two pivots (in the first and third columns) the three columns must be linearly dependent, and in fact the second column is twice the first. We can therefore remove the second column to form the following 2 by 2 matrix:

$A'' = \begin{bmatrix} 1&1 \\ 0&4 \end{bmatrix}$

Since $A''$ is in echelon form and has pivots in all columns the columns are linearly independent. They also span $\mathbf{R}^2$ with any vector $w = (w_1, w_2)$ expressible as

$w = (w_1 - \frac{1}{4}w_2) \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \frac{1}{4}w_2 \begin{bmatrix} 1 \\ 4 \end{bmatrix}$

Since the two columns of $A''$ are linearly independent and span $\mathbf{R}^2$ they form a basis for $\mathbf{R}^2$.

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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