Linear Algebra and Its Applications, Exercise 3.4.3

Posted on December 16, 2017 by hecker

Exercise 3.4.3. Given the 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) from the previous exercise, project b onto a third orthonormal vector a_3 = (\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}). What is the sum of the three projections? Why? Why is the matrix P = a_1a_1^T + a_2a_2^T + a_3a_3^T equal to the identity matrix I?

Answer: Since a_3 is orthonormal, the projection of b onto a_3 is given by

a_3^Tba_3 = \begin{bmatrix} \frac{2}{3}&-\frac{1}{3}&\frac{2}{3} \end{bmatrix} \begin{bmatrix} 0 \\ 3 \\ 0 \end{bmatrix} \begin{bmatrix} \frac{2}{3} \\ -\frac{1}{3} \\ \frac{2}{3} \end{bmatrix} = -1 \cdot \begin{bmatrix} \frac{2}{3} \\ -\frac{1}{3} \\ \frac{2}{3} \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} \\ \frac{1}{3} \\ -\frac{2}{3} \end{bmatrix}

The sum of all three projections is then:

a_1^Tba_1 + a_2^Tba_2 + a_3^Tba_3 = \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{1}{3} \\ -\frac{2}{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 0 \end{bmatrix} = b

This is because the vectors a_1, a_2, and a_3 together form an orthonormal basis for \mathbb{R}^3, and any vector v in \mathbb{R}^3 can be expressed as a linear combination of those basis vectors with coefficients given by the projection of v onto each basis vector.

We then have

b = a_1^Tba_1 + a_2^Tba_2 + a_3^Tba_3 = a_1a_1^Tb + a_2a_2^Tb + a_3a_3^Tb = (a_1a_1^T + a_2a_2^T + a_3a_3^T)b

(taking advantage of the fact that a_1^Tb, etc., are scalars) so that the 3 by 3 matrix

P = a_1a_1^T + a_2a_2^T + a_3a_3^T = I

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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