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Let $T$ be a linear transformation on $\mathbb{R}^{4}$ whose standard matrix is $$\left(\begin{array}{rrrr} 1 & -1 & -1 & -1\\ 1 & 1 & 1 & -1\\ 1 & 1 & -1 & 1\\ 0 & 0 & 0 & 0 \end{array}\right).$$ Does there exist a $3$-dimensional subspace of $\mathbb{R}^{4}$, $V$, and a linear transformation $S$ on $\mathbb{R}^{4}$ such that $S(\mathbf{v})=T(\mathbf{v})$ for all $\mathbf{v}\in V$ and $\left\Vert S(\mathbf{x})\right\Vert _{1}\le2$ for all $\mathbf{x}\in\mathbb{R}^{4}$ with $\left\Vert \mathbf{x}\right\Vert _{1}=1$?

Thanks in advance for any helpful answers.

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If this is a homework problem, please use the homework tag. What have you tried so far? –  Adam Saltz Nov 1 '11 at 17:25
    
Since $\|S(kx)\|_1 = \|kS(x)\|_1 = |k| \|S(x)\|_1$, I rather doubt it... –  user7530 Nov 1 '11 at 17:26
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Please stop deleting questions. –  Mariano Suárez-Alvarez Nov 5 '11 at 19:54
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1 Answer 1

Let $(1,1,0,0),(0,1,0,1),(0,0,1,1)$ be a basis for your subspace. And let

$S = \left(\begin{matrix} 0 & 0 & 0 & -2 \\ 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0\end{matrix}\right).$

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@189: Mostly trial and error. It helped to notice that the subspace can't contain any of the standard basis vectors $e_1,e_2,e_3,e_4$. I first tried $(1,1,0,0)$, $(0,1,1,0)$, $(0,0,1,1)$, but that didn't work. Then I saw that $(0,0,1,1)$ is probably good because I can get that $2\times2$ block of zeros at the right part of $S$. Anyway, I understand your concern and I'd be happy myself to see a more systematic way of finding such a subspace. –  J. J. Nov 1 '11 at 18:56
    
@J.J. How would one go about proving that such a subspace doesn't exist (if in fact it didn't)? –  Weltschmerz Nov 1 '11 at 21:32
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@Weltschmerz: sadly I don't know. Anyway, It seems to me that if we were using the Euclidean norm (which is induced by the inner product) instead of the Manhattan norm, the question might become easier. I suspect that then these subspaces are more tractable by looking at the eigenvectors of $T$. This is because if $v_1,\dots,v_k$ is an orthonormal basis for $V$, then we can extend it into an orthonormal basis for the whole space and define $S$ by $S(v_j) = T(v_j)$ for $j=1,\dots,k$ and $S(v) = 0$ for the other basis vectors. The norm of $S$ will be the norm of $T|V$. –  J. J. Nov 2 '11 at 6:07
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