Linear Algebra and Its Applications, Exercise 3.1.16

Exercise 3.1.16. Describe the set of all vectors orthogonal to the vectors (1, 4, 4, 1) and (2, 9, 8, 2).

Answer: If a vector x is orthogonal to the vectors (1, 4, 4, 1) and (2, 9, 8, 2) then its inner products with those vectors must be zero, so that

1 \cdot x_1 + 4 \cdot x_2 + 4 \cdot x_3 + 1 \cdot x_4 = 0

and

2 \cdot x_1 + 9 \cdot x_2 + 8 \cdot x_3 + 2 \cdot x_4 = 0

This is a system of two equations in four unknowns:

\begin{array}{rcrcrcrcl} x_1&+&4x_2&+&4x_3&+&x_4&=&0 \\ 2x_1&+&9x_2&+&8x_3&+&2x_4&=&0 \end{array}

and is equivalent to the system Ax = 0 where

A = \begin{bmatrix} 1&4&4&1 \\ 2&9&8&2 \end{bmatrix}

In other words, the problem of finding all vectors orthogonal to (1, 4, 4, 1) and (2, 9, 8, 2) is equivalent to the problem of finding the nullspace of the 2 by 4 matrix A for which (1, 4, 4, 1) and (2, 9, 8, 2) are the rows.

We use Gaussian elimination to solve the system, starting by subtracting 2 times row 1 from row 2:

\begin{bmatrix} 1&4&4&1 \\ 2&9&8&2 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&4&4&1 \\ 0&1&0&0 \end{bmatrix}

The resulting echelon matrix has two pivots, in columns 1 and 2, so x_1 and x_2 are basic variables and x_3 and x_4 are free variables.

Setting x_3 = 1 and x_4 = 0, from row 2 we have x_2 = 0 and from row 1 we then have x_1 + 4x_2 + 4x_3 + x_4 = x_1 + 0 + 4 + 0 = 0 or x_1 = -4. So (-4, 0, 1, 0) is one solution to Ax = 0. Setting x_3 = 0 and x_4 = 1, from row 2 we have x_2 = 0 again and from row 1 we have x_1 + 4x_2 + 4x_3 + x_4 = x_1 + 0 + 0 + 1 = 0 or x_1 = -1. So (-1, 0, 0, 1) is a second solution to Ax = 0.

The vectors (-4, 0, 1, 0) and (-1, 0, 0, 1) together form a basis for the nullspace of A. That nullspace, i.e., all linear combinations of (-4, 0, 1, 0) and (-1, 0, 0, 1), is the set of all vectors orthogonal to the vectors (1, 4, 4, 1) and (2, 9, 8, 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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