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Given a vector $x \in \mathbb{R}^n$ we know the following inequality holds for the product of the vector $x$ and a matrix $A \in \mathbb{R}^{m\times n}$ i.e., $Ax=y$ where $y \in \mathbb{R}^m$

$\Vert y\Vert_2=\Vert Ax\Vert_2\leq \Vert A\Vert_F\Vert x\Vert_2$

1) Can we say $x$ is linearly independent of rows of $A$ when inequality (<) holds.

or in other words

1) Can we say $x$ is linearly dependent on rows of $A$ when equality (=) holds.

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We have $=$ if and only if $x$ and each row of $A$ are linearly dependent. This follows directly from Cauchy Schwarz inequality.

Proof: Let $a_i^T$ denote the $i$-th row of $A$. Then, we have $$ \| Ax \|^2 = \sum_{i=1}^m (a_i^T x)^2 \le \sum_{i=1}^m \|a_i\|^2 \|x\|^2 = \|A\|_{\mathrm F}^2 \| x \|^2, $$ where the inequality follows from Cauchy Schwarz. Now, by Cauchy Schwarz, we have $(a_i^T x)^2 = \|a_i\|^2 \|x\|^2$ if and only if $a_i$ and $x$ are linearly dependent.

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  • $\begingroup$ Can you help in a simple proof. $\endgroup$ – Astro Jul 15 '15 at 13:38

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