## 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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