Linear Algebra and Its Applications, Exercise 3.4.8

Posted on December 21, 2017 by hecker

Exercise 3.4.8. Project the vector b = (1, 2) onto the two non-orthogonal vectors a_1 = (1, 0) and a_2 = (1, 1) and show that the sum of the two projections does not equal b (as it would if a_1 and a_2 were orthogonal).

Answer: The projection of b onto a_1 is (a_1^Tb/a_1^Ta_1)a_1. We have a_1^Tb = (1 \cdot 1 + 0 \cdot 2) = 1 and a_1^Ta_1 = (1 \cdot 1 + 0 \cdot 0) = 1. So the projection of b onto a_1 is \frac{1}{1} a_1 = (1, 0).

Similarly, the projection of b onto a_2 is (a_2^Tb/a_2^Ta_2)a_2. We have a_2^Tb = (1 \cdot 1 + 1 \cdot 2) = 3 and a_2^Ta_2 = (1 \cdot 1 + 1 \cdot 1) = 2. So the projection of b onto a_2 is \frac{3}{2} a_2 = (\frac{3}{2}, \frac{3}{2}).

The sum of the two projections is (1, 0) + (\frac{3}{2}, \frac{3}{2}) = (\frac{5}{2}, \frac{3}{2}), which is not equal to b.

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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