## 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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