Linear Algebra and Its Applications, Exercise 3.2.11

Exercise 3.2.11. a) Given the line through the origin and a = \left( 1, 3 \right) find the matrix P_1 that projects onto this line, as well as the matrix P_2 that projects onto the line perpendicular to the original line.

b) What is P_1 + P_2? What is P_1P_2? Explain your answers.

Answer: a) We have

P_1 = aa^T/a^Ta = \frac{1}{1^2 + 3^2} \begin{bmatrix} 1 \\ 3 \end{bmatrix}\begin{bmatrix} 1&3 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 1&3 \\ 3&9 \end{bmatrix}

We next find the perpendicular line. If this line passes through the origin and b = \left(b_1, b_2 \right) then we will have b orthogonal to a so that a^Tb = a_1b_1 + a_2b_2 = b_1 + 3b_2 = 0. One vector satisfying this condition is b = \left( -3, 1 \right). We then have

P_2 = bb^T/b^Tb = \frac{1}{(-3)^2 + 1^2} \begin{bmatrix} -3 \\ 1 \end{bmatrix}\begin{bmatrix} -3&1 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 9&-3 \\ -3&1 \end{bmatrix}

b) Taking the product of the two projection matrices first, we have

P_1P_2 = \frac{1}{10} \begin{bmatrix} 1&3 \\ 3&9 \end{bmatrix} \frac{1}{10} \begin{bmatrix} 9&-3 \\ -3&1 \end{bmatrix}

= \frac{1}{100} \begin{bmatrix} 9-9&-3+3 \\ 27-27&-9+9 \end{bmatrix} = \frac{1}{100} \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} = 0

This true because multiplying a vector by P_2 projects it onto the line through b, and multiplying by P_1 projects the resulting vector onto the line through a. But because b is perpendicular to a in multiplying by P_1 all points on the line through b get projected onto the origin.  The effect of multiplying by the product matrix P_1P_2 is thus the same as multiplying by zero. (Note also that P_2P_1 = 0 for the same reason.)

Taking the sum of the two projection matrices, we have

P_1 + P_2 = \frac{1}{10} \begin{bmatrix} 1&3 \\ 3&9 \end{bmatrix} + \frac{1}{10} \begin{bmatrix} 9&-3 \\ -3&1 \end{bmatrix}

= \frac{1}{10} \begin{bmatrix} 1+9&3-3 \\ 3-3&9+1 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 10&0 \\ 0&10 \end{bmatrix}

= \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I

In other words, multiplying a vector by the sum of the two projection matrices results in the same vector:

(P_1+P_2)v = P_1v + P_2v = v

In effect we have transformed the vector v into two perpendicular component vectors: one component P_1v on the line through a and a second component P_2v on the perpendicular line through b. Adding the components back together reconstitutes the original vector.

UPDATE: Corrected the calculations above to include the factors 1/a^Ta and 1/b^Tb when calculating the projection matrices. Thanks go to Trystyn for catching my original error.

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

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2 Responses to Linear Algebra and Its Applications, Exercise 3.2.11

  1. Trystyn says:

    This is wrong.
    P = [a a^T] / [a^T a]
    not just [a a^T]

  2. hecker says:

    Trystyn: You are correct, thanks for finding this error! I’ve updated the post to reflect your correction.

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