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Consider the following subspaces of $\mathbb R^3$ :

$U=\{\ (x,y,z) \in \mathbb R^3:2x-y+z=0\}$ and

$W=\langle\{(2,-1,0),(1,0,1,) \}\rangle$

2.1 )Find a basis for $U+W$

What i've done \begin{align} U& =\{\ (x,y,z) \in \mathbb R^3:2x-y+z=0\} \\ & = \{\ (x,y,z) \in \mathbb R^3:z=-2x+y\}\\ & = \{\ (x,y,-2x+y)\}\\ & = (x,0,-2x)+(0,y,y)\\ & = x(1,0,-2)+y(0,1,1)=(0,0,0)\\ & = \left\{ \begin{array}{c} x=0 \\ y=0 \\ -2x+y=0 \end{array} \right. \\ \end{align} Thus, U generator is $\langle (1,0,-2),(0,1,1)\rangle$ is lineal independent and Basis of U, with both scalars $x,y=0$

And the Basis of U+W , for property of span, we can say :

$U+W=\langle U \cup W \rangle$ $\implies$ $\langle (1,0,-2);(0,1,1);(2,-1,0);(1,0,1)\rangle$ And then how to determine the dimension?, and lineal independence?

For this example i have been doing the matrix, and the result were $\infty$ solutions, because range of the matrix and R(a/b) is still 3, and number of unknowns are 4, what does it mean?

2.2)Characterice W

In this example i've found the basis of W, and using matrices and RREF, that brought me to a matrix with unique vectors

$$ \left[ \begin{array}{cc|c} 2&1&0\\ 0&0&0 \\ 0&0&0 \end{array} \right] $$

and that means W is a basis and has dim(2)

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2 Answers 2

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You are right that $U$ is the span of $(1,0,-2)$ and $(0, 1, 1)$ (although some of the intermediate steps you have written to get there do not make sense...), and that $U+W$ is the span of $(1,0,-2)$, $(0, 1,1)$, $(2,-1,0)$, and $(1,0,1)$.

If you put the above four vectors into a $4 \times 3$ matrix and find the RREF, you can obtain a basis for $U+W$.

I'm not sure how to answer your last question since you haven't defined $V$.

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  • $\begingroup$ My mistake!, it was W and not V. How can I force the basis of $U+W$?, I mean, by deleting a column? $\endgroup$ Commented Apr 27, 2019 at 5:31
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We see U is the eq of a plane, dimension 2. W not equal U. Both are subspace of R3, dimension 3 thus any basis of R3 will do.

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