Linear Algebra and Its Applications, Review Exercise 2.1

Review exercise 2.1. For each of the following subspaces of \mathbb{R}^4 find a suitable set of basis vectors:

a) all vectors for which x_1 = 2x_4

b) all vectors for which x_1+x_2+x_3 = 0 and x_3+x_4=0

c) all vectors consisting of linear combinations of the vectors (1, 1, 1, 1), (1, 2, 3, 4), and (2, 3, 4, 5)

Answer: a) Since there are no constraints on x_2, x_3, and x_4 they may assume any value. (In other words, x_2, x_3, and x_4 are all free variables and only x_1 is a basic variable.) If x_2 = x_3 = 0 and x_4 = 1 then the  vector (2, 0, 0, 1) is in the subspace and can serve as a basis vector. If x_2 = 1 and x_3 = x_4 = 0 then we obtain a second basis vector (0, 1, 0, 0) and if x_2 = x_4 = 0 and x_3 = 1 then we obtain a third basis vector (0, 0, 1, 0).

The vector (2, 0, 0, 1), (0, 1, 0, 0) and (0, 0, 1, 0) thus form a basis for the subspace.

b) The two equations form the linear system

\begin{bmatrix} 1&1&1&0 \\ 0&0&1&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = 0

In this system x_1 and x_3 are basic variables and x_2 and x_4 are free variables. If we set x_2 = 1 and x_4 = 0 then we have x_3 = 0 and x_1 = -1. If we set x_2 = 0 and x_4 = 1 then we have x_3 = -1 and x_1 = 1.

The two vectors (-1, 1, 0, 0) and (1, 0, -1, 1) are thus basis vectors for the subspace (which happens to be the nullspace of the matrix above).

c) We have (2, 3, 4, 5) = (1, 1, 1, 1) + (1, 2, 3, 4). So of the three vectors only two are linearly independent, and (1, 1, 1, 1) and (1, 2, 3, 4) can serve as a basis for the subspace spanned by the vectors.

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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