## Linear Algebra and Its Applications, Exercise 3.4.7

Exercise 3.4.7. Given $b = x_1q_1 + x_2q_2 + \cdots + x_nq_n$ where $q_1, q_2, \dots, q_n$ are orthonormal vectors, compute $b^Tb$ and show that

$\|b\|^2 = x_1^2 + x_2^2 + \cdots + x_n^2$

Answer:We have $b = \sum_i x_iq_i$ so that

$b^Tb = (\sum_i x_iq_i)^T(\sum_j x_jq_j) = (\sum_i x_iq_i^T)(\sum_j x_jq_j)$

since the transpose of a sum is equal to the sum of the transposes. The product of the sums can then be decomposed into two sums of products as follows:

$(\sum_i x_iq_i^T)(\sum_j x_jq_j) = \sum_{i=j} (x_iq_i^T)(x_jq_j) + \sum_{i \ne j} (x_iq_i^T)(x_jq_j)$

$= \sum_{i=j} x_ix_jq_i^Tq_j + \sum_{i \ne j} x_ix_jq_i^Tq_j$

But since $q_1, q_2, \dots, q_n$ are orthonormal vectors we have $q_i^Tq_j = 1$ when $i = j$ and $q_i^Tq_j = 0$ when $i \ne j$, so that

$\sum_{i=j} x_ix_jq_i^Tq_j + \sum_{i \ne j} x_ix_jq_i^Tq_j = \sum_{i = j} x_ix_j \cdot 1 + \sum_{i \ne j} x_ix_j \cdot 0 = \sum_i x_i^2$

We thus have

$\|b\|^2 = \sum_i x_i^2 = x_1^2 + x_2^2 + \cdots + x_n^2$

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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