## Linear Algebra and Its Applications, Exercise 3.4.8

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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