Norm of a particular matrix (regarding inequality)

Question

Prove that $\|aa^T\| \le \|a\|^2$ where $a\in \mathbb{R}^n$. The norm of a vector here is the Euclidean norm and norm of a matrix $A$ is $\|A\|=\max\{\|Ax\|:\|x\|=1\}$ and if $A$ is symmetric then $\|A\|=\max\{|x^TAx|:\|x\|=1\}$.

Answer 1

Using the definition of the matrix norm for symmetric matrices stated in the question, you have that

$$ \|aa^T\| = \max_{\|x\|=1} |x^Taa^Tx|. $$

Given that $|x^Taa^Tx| = |x^Ta|^2$, it follows by the Cauchy-Schwarz inequality that

$$ |x^Taa^Tx| = |x^Ta|^2 \leq \|x\|^2\|a\|^2 \leq \|a\|^2 $$

for all $x\in\mathbb{R}^n$ with $\|x\| = 1$. Thus, $\|aa^T\| \leq \|a\|^2$.

Answer 2

Since $aa^T$ is symmetric, we have $\|aa^T\|=\max \{|x^Taa^Tx|:\|x\|=1\}$. Let $x=(\frac{x_1}{N},\cdots,\frac{x_1}{N})$ where $N=\sqrt(\sum x_i^2)$ so that $\|x\|=1$. So $x^Ta= \frac{\sum x_ia_i}{N}$. Hence $x^Taa^Tx= (\frac{\sum x_ia_i}{N})^2$. So we need to prove that $(\frac{\sum x_ia_i}{N})^2 \le \|a\|^2=\sum a_i^2$ i.e $(\sum x_ia_i)^2 \le \sum x_i^2\sum a_i^2$.

Proof by induction on $n$. The statement is obvious for $n=1$, assume true for $n$. Now the left side of the inequality for $n+1$ is $(\sum^{n+1}x_ia_i)^2 \leq \sum^nx_i^2 \sum^n a_i^2 + 2(\sum^nx_ia_i)x_{n+1}a_{n+1}+x_{n+1}^2a_{n+1}^2$ (using $(a+b)^2=a^2+2ab+b^2$ and the induction hypothesis). Call this expression $A$. The right side of the inequality for $n+1$ is $\sum^nx_i^2\sum^na_i^2+x_{n+1}\sum^na_i^2+a_{n+1}\sum^nx_i^2+x_{n+1}^2a_{n+1}^2$. Call this expression $B$. Now we have to prove that $B-A\ge 0$.

\begin{equation} B-A = (a_1x_{n+1})^2+(a_2x_{n+1})^2+\cdots+(a_nx_{n+1})^2+(a_{n+1}x_1)^2+(a_{n+1}x_2)^2+\cdots +(a_{n+1}x_n)^2 - 2x_1a_1x_{n+1}a_{n+1}- 2x_2a_2x_{n+1}a_{n+1}\cdots - 2x_na_nx_{n+1}a_{n+1} \end{equation} which can be written as a sum of squares, hence $B-A\geq 0$.