Linear Algebra and Its Applications, Exercise 3.4.28

Exercise 3.4.28. Given the plane x_1 + x_2 + x_3 = 0 and the following vectors

\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}

in the plane, find an orthonormal basis for the subspace represented by the plane. Report the dimension of the subspace and the number of nonzero vectors produced by Gram-Schmidt orthogonalization.

Answer: We start with the vector a_1 = (1, -1, 0) and normalize it to create q_1:

\|a_1\|^2 = 1^2 + (-1)^2 + 0^2 = 1 + 1 = 2

q_1 = a_1/\|a_1\| = \frac{1}{\sqrt{2}} a_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix}

We then take the second vector a_2 = (0, 1, -1) and create a second orthogonal vector a_2' by subtracting from a_2 its projection on q_1:

a_2' = a_2 - (q_1^Ta_2)q_1

= a_2 - \left[ \frac{1}{\sqrt{2}} \cdot 0 + (-\frac{1}{\sqrt{2}}) \cdot 1 + 0 \cdot (-1) \right]q_1 = a_2 - (-\frac{1}{\sqrt{2}})q_1 = a_2 + \frac{1}{\sqrt{2}}q_1

= \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + \frac{1}{\sqrt{2}} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + \begin{bmatrix} \frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ -1 \end{bmatrix}

We then normalize a_2' to create q_2:

\|a_2'\|^2 = (\frac{1}{2})^2 + (\frac{1}{2})^2 + (-1)^2 = \frac{1}{4} + \frac{1}{4} + 1 = \frac{3}{2}

q_2 = a_2'/\|a_2'\| = a_2'/\sqrt{\frac{3}{2}} = \frac{\sqrt{2}}{\sqrt{3}} \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ -1 \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{2}}{2\sqrt{3}} \\ \frac{\sqrt{2}}{2\sqrt{3}} \\ -\frac{\sqrt{2}}{\sqrt{3}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \end{bmatrix}

Finally, we take the third vector a_3 = (1, 0, -1) and attempt to create another orthogonal vector a_3' by subtracting from a_3 its projections on q_1 and q_2:

a_3' = a_3 - (q_1^Ta_3)q_1 - (q_2^Ta_3)q_2

= a_3 - \left[ \frac{1}{\sqrt{2}} \cdot 1 + (-\frac{1}{\sqrt{2}}) \cdot 0 + 0 \cdot (-1) \right]q_1- \left[ \frac{1}{\sqrt{6}} \cdot 1 + \frac{1}{\sqrt{6}} \cdot 0 + (-\frac{2}{\sqrt{6}}) \cdot (-1) \right] q_2

= a_3 - \frac{1}{\sqrt{2}}q_1 - \frac{3}{\sqrt{6}}q_2 = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} - \frac{1}{\sqrt{2}} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} - \frac{3}{\sqrt{6}} \begin{bmatrix} \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \end{bmatrix}

= \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} - \begin{bmatrix} \frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix} - \begin{bmatrix} \frac{3}{6} \\ \frac{3}{6} \\ -\frac{6}{6} \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} - \begin{bmatrix} \frac{1}{2} \\ -\frac{1}{2} \\ 0 \end{bmatrix} - \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

Since a_3' = 0 we cannot create a third orthogonal vector to q_1 and q_2. The vectors

q_1 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} \qquad q_2 = \begin{bmatrix} \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \end{bmatrix}

are an orthonormal basis for the subspace, and the dimension of the subspace is 2.

(In hindsight we could have predicted this result by inspecting the original vectors a_1, a_2, and a_3 and noticing that a_3 = a_1 + a_2. Thus only a_1 and a_2 were linearly independent, a_3 being linearly dependent on the first two vectors, so that only two orthonormal basis vectors could be created from the three vectors given.)

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, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

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