## Linear Algebra and Its Applications, Exercise 3.4.3

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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