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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