Hey all, another subspace related question..

how can I show that if $\displaystyle P$ is a subspace of $\displaystyle R^m$, then $\displaystyle U = \{\vec{u} \in R^n: A\vec{u} \in P\}$ is a subspace of R^n, where A is an m*n matrix.

Thanks.

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- Aug 4th 2011, 04:18 PMOilershow that P is a subspace..
Hey all, another subspace related question..

how can I show that if $\displaystyle P$ is a subspace of $\displaystyle R^m$, then $\displaystyle U = \{\vec{u} \in R^n: A\vec{u} \in P\}$ is a subspace of R^n, where A is an m*n matrix.

Thanks. - Aug 4th 2011, 07:16 PMFernandoRevillaRe: show that P is a subspace..
(i) $\displaystyle A0=0\in P$ (because $\displaystyle P$ is subspace). Hence $\displaystyle 0\in U$.

(ii) For all $\displaystyle u,v\in U$ we have $\displaystyle Au\in P$ and $\displaystyle Av\in P$. Then $\displaystyle A(u+v)=Au+Av\in P$ (because $\displaystyle P$ is subspace) . Hence $\displaystyle u+v\in U$ .

(iii) ... Try it.