## Linear Algebra and Its Applications, Exercise 2.3.9

Exercise 2.3.9. Give a basis for the column space of the matrix $U = \begin{bmatrix} 0&1&4&3 \\ 0&0&2&2 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix}$

and express the other columns of $U$ in terms of it. Find a matrix $A$ that is reduced by elimination to the same $U$ but has a different column space than $U$.

Answer: The pivots of $U$ are in the second and third columns, so that there columns are linearly independent and can serve as a basis for the column space. We can express the fourth column as a linear combination of the second and third columns as follows: $\begin{bmatrix} 3 \\ 2 \\ 0 \\ 0 \end{bmatrix} = - \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 4 \\ 2 \\ 0 \\ 0 \end{bmatrix}$

We can construct a matrix $A$ for which elimination produces $U$ simply by having the first two rows of $A$ be the same as those of $U$ and then making the third row of $A$ equal to the second row of $U$ instead of being set to zeros: $A = \begin{bmatrix} 0&1&4&3 \\ 0&0&2&2 \\ 0&0&2&2 \\ 0&0&0&0 \end{bmatrix}$

The column space of $A$ is spanned by the basis vectors $(1, 0, 0, 0)$ and $(4, 2, 2)$ and is different than the column space of $U$.

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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