Let $A \in \mathbb{R}^{m\times n}$ and $v, w \in \mathbb{R}^n$, $u\in \mathbb{R}^m$, where $u,v,w$ are all unit vectors. Prove of disprove that $\max |Av| = \max u\cdot (Aw)$. (Assuming $|\cdot|$ is vector 2-norm)
Prove or disprove: maximum of magnitude of $|Av|$ is equal to maximum of $u \cdot (Aw)$
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linear-algebra
2 Answers
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If $A \in \mathbb{R^{m \times n}}$ then we can make a singular value decomposition $A = U \Sigma V^T$. We can choose the row vector $v^T$ from $V^T$ that corresponds to the largest value of $\Sigma$ and the column vector $u$ from $U$ that corresponds to the same value.
now $\max\limits_{|r| = 1} |A r| = |A v| = u^TAv = \max\limits_{|q| = 1, |r|=1} qAr$
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Let $A=\pmatrix{-1&-2\\-3&-4}$ then $$\max |A v|=\max\{2,4\}=4$$ But $$\max u\cdot(Aw)=\max_{i,j}A_{ij}=-1$$
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0BTW, I assumed that By $|\cdot|$ you meant **norm-1** and by _unit vactor_ you meant a vector that has only one nonzero element with value $1$ – 2017-02-01
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0Actually I mean norm-2. – 2017-02-01
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0Either way the statement is false @Lewis – 2017-02-01