# Is $\left \{ x_{1}+y_{1},..., x_{n}+y_{n}\right \}$ a basis for $\mathbb{R}^{n}$?

Suppose $\left \{ x_{1},..., x_{n}\right \}$ and $\left \{ y_{1},..., y_{n}\right \}$ are two different bases for $\mathbb{R}^{n}$. Is $\left \{ x_{1}+y_{1},..., x_{n}+y_{n}\right \}$ also a basis for $\mathbb{R}^{n}?$ If yes, prove your claim, if no, show a counterexample.

I think the claim is true. I have picked two different bases in $\mathbb{R}^{2}$ to try to come up with a counterexample, but couldn't. I picked $\left \{ (0,1),(1,0) \right \}$ and $\left \{ (1,1),(-1,1) \right \}$ and added their components to come up with a different set of vectors $\left \{ (1,2),(0,1) \right \}$. However, I have found that this is also a basis for $\mathbb{R}^{2}$ as it spans $\mathbb{R}^{2}$ and also is linearly independent, that is, $Ax=b \Rightarrow x=0$.

If the claim is true, then can anyone help me with the proof?

• What of $y_i = -x_i$? Sep 2, 2015 at 9:28

If you chose $y_i=-x_i$ the claim is false.