## 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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