# Find a basis of the following subspace in $\mathbb R^3$

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)

## 2 Answers

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 $$V$$.

• My mistake!, it was W and not V. How can I force the basis of $U+W$?, I mean, by deleting a column? Commented Apr 27, 2019 at 5:31

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.