Please explain what results we use to get the following:
$\|AA^T(Ax-b)\| \leq \|A^TA\|\cdot\|(Ax-b)\|,$
where $A$ is an $m \times n$ matrix; $b$ and $x$ are $m \times 1$ vectors, and $\|.\|$ is the $2$-norm.
Please explain what results we use to get the following:
$\|AA^T(Ax-b)\| \leq \|A^TA\|\cdot\|(Ax-b)\|,$
where $A$ is an $m \times n$ matrix; $b$ and $x$ are $m \times 1$ vectors, and $\|.\|$ is the $2$-norm.
For any matrix $A$ and vector $b$, using Cauchy-Schwarz in the second inequality, $ \|Ab\|_2^2=\sum_j\left|\sum_kA_{jk}b_k\right|^2\leq\sum_j\left(\sum_k|A_{jk}b_k|\right)^2 \leq\sum_j\left(\left(\sum_k|A_{jk}|^2\right)^{1/2}\left(\sum_k|b_k|^2\right)^{1/2}\right)^2 =\sum_j\sum_k|A_{jk}|^2\sum_k|b_k|^2=\|A\|_2^2\|b\|_2^2. $ So $\|Ab\|_2\leq\|A\|_2\|b\|_2$. Now you can use $AA^T$ instead of $A$ and $Ax-b$ instead of $b$.