## Linear Algebra and Its Applications, Exercise 3.2.7

Exercise 3.2.7. Show that $\left(a_1 + \dots + a_n\right)^2 \le n\left(a_1^2 + \dots + a_n^2\right)$. Hint: use the Schwarz inequality $|a^Tb| \le \|a\|\|b\|$ with an appropriate choice of $b$.

Answer: As noted in the hint, the key to proving this is to find an appropriate choice of $b$. The easiest way to do this is to work backwards from the statement $\left(a_1 + \dots + a_n\right)^2 \le n\left(a_1^2 + \dots + a_n^2\right)$ that we want to prove.

First, note that $\left(a_1^2 + \dots + a_n^2\right) = \|a\|^2$. Since the expression $n\left(a_1^2 + \dots + a_n^2\right)$ is on the right  side of the inequality we wish to prove, and the expression $\|a\|\|b\| = \|b\|\|a\|$ is on the right side of the Schwarz inequality, this suggests that we should choose $b$ so that $n\left(a_1^2 + \dots + a_n^2\right) = \|b\|^2\|a\|^2$, or $n = \|b\|^2$ .

Second, since the expression $\|b\|^2 \|a\|^2 = \left(\|a\|\|b\|\right)^2$ is the square of the expression on the right side of the Schwarz inequality, this suggests that we should choose $b$ so that$\left(a_1+\dots+a_n\right)^2$ is the square of the expression $|a^Tb|$ on the left side of the Schwarz inequality, so that $|a^Tb| = |a_1+\dots+a_n|$. Since $a^Tb = a_1b_1 + \dots + a_nb_n$ this suggests that we choose $b = \left(1, \dots, 1\right)$.

With this choice of $b$ the proof is simple. If we take the Schwarz inequality $|a^Tb| \le \|a\|\|b\|$ and square both sides we obtain $|a^Tb|^2 \le \left(\|a\|\|b\|\right)^2$. If $b = \left(1, \dots, 1\right)$ then $a^Tb = a_1+\dots+a_n$ as noted earlier. Whether $a^Tb$ is positive or negative squaring it produces the same result, so that $|a^Tb|^2 = \left(a^Tb\right)^2 = \left(a_1+\dots+a_n\right)^2$ on the left hand side of the inequality.

If If $b = \left(1, \dots, 1\right)$ then $\|b\|^2 = \sum_{i=1}^n 1^2 = n$ so that we have $\left(\|a\|\|b\|\right)^2 = \|b\|^2\|a\|^2 = n\left(a_1^2+\dots+a_n^2\right)$ on the right hand side of the inequality.

Combining the left and right sides of the inequality we then have $\left(a_1 + \dots + a_n\right)^2 \le n\left(a_1^2 + \dots + a_n^2\right)$.

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, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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