Linear Algebra and Its Applications, Exercise 2.1.8

Exercise 2.1.8. Consider the following system of linear equations:

Ax = \begin{bmatrix} 1&1&1 \\ 1&0&2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

Does the set of solutions x form a point, line, or plane? Is it a subspace? Is it the nullspace of A? The column space of A?

Answer: The system Ax = 0 corresponds to

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}x_1&+&x_2&+&x_3&=&0 \\ x_1&&&+&2x_3&=&0 \end{array}

From the second equation we have x_1 = -2x_3 and we can substitute into the first equation to obtain -2x_3 + x_2 + x_3 = 0 or x_2 - x_3 = 0. We therefore have x_2 = x_3 so that the set of solutions x can be represented as

x = \begin{bmatrix} -2x_3 \\ x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix}

where x_3 is a free variable. The set of solutions x is therefore a line passing through the origin and the point (-2, 1, 1).

Since the solution set is a line passing through the origin it is a subspace, and since x is the set of all vectors for which Ax = 0 it is by definition the nullspace \mathcal{N}(A) of A (see page 68).

The column space of A is the set of all vectors v that are linear combinations of the columns of A:

v = c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 2 \end{bmatrix}

and thus contains 2 by 1 vectors. All solution vectors x are 3 by 1 vectors; the set of solutions x is not the same as the column space of A.

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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