## 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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