Linear Algebra and Its Applications, Exercise 3.4.2

Posted on December 15, 2017 by hecker

Exercise 3.4.2. Given two orthonormal vectors a_1 = (\frac{2}{3}, \frac{2}{3}, -\frac{1}{3}) and a_2 = (-\frac{1}{3}, \frac{2}{3}, \frac{2}{3}) and the vector b = (0, 3, 0), project b onto a_1 and a_2. Also find the projection p of b onto the plane formed by a_1 and a_2.

Answer: Since a_1 is orthonormal, the projection of b onto a_1 is given by

a_1^Tba_1 = \begin{bmatrix} \frac{2}{3}&\frac{2}{3}&-\frac{1}{3} \end{bmatrix} \begin{bmatrix} 0 \\ 3 \\ 0 \end{bmatrix} \begin{bmatrix} \frac{2}{3} \\ \frac{2}{3} \\ -\frac{1}{3} \end{bmatrix} = 2 \cdot \begin{bmatrix} \frac{2}{3} \\ \frac{2}{3} \\ -\frac{1}{3} \end{bmatrix} = \begin{bmatrix} \frac{4}{3} \\ \frac{4}{3} \\ -\frac{2}{3} \end{bmatrix}

Similarly the projection of b onto a_2 is given by

a_2^Tba_2 = \begin{bmatrix} -\frac{1}{3}&\frac{2}{3}&\frac{2}{3} \end{bmatrix} \begin{bmatrix} 0 \\ 3 \\ 0 \end{bmatrix} \begin{bmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix} = 2 \cdot \begin{bmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} \\ \frac{4}{3} \\ \frac{4}{3} \end{bmatrix}

The projection p of b onto the plane formed by a_1 and a_2 is simply the sum of the projections above:

p = a_1^Tba_1 + a_2^Tba_2 = \begin{bmatrix} \frac{4}{3} \\ \frac{4}{3} \\ -\frac{2}{3} \end{bmatrix} + \begin{bmatrix} -\frac{2}{3} \\ \frac{4}{3} \\ \frac{4}{3} \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ \frac{8}{3} \\ \frac{2}{3} \end{bmatrix}

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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