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You are given two vector subspaces U = span $\left(\begin{bmatrix} 1 \cr 1 \cr 1 \cr 1 \cr 1\end{bmatrix} ,\begin{bmatrix} 1 \cr 0 \cr 0 \cr 0 \cr-1 \end{bmatrix} \right)$ and V = span $\left(\begin{bmatrix} -1 \cr -1 \cr0 \cr0\cr1 \end{bmatrix}, \begin{bmatrix} 3 \cr 2 \cr2 \cr 2\cr1 \end{bmatrix}\begin{bmatrix} 2 \cr 1\cr2\cr2\cr2 \end{bmatrix} \right)$.

How would you find the basis of U+V and the basis of U∩V?

I have problems with the linearly dependent vector in the span of V. When am I able to remove this vector and perform Gauss elimination of a 5x4 matrix instead of the whole 5x5 matrix (for the sum or intersection?)

Vae
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  • One would apply the standard way, see for example here for the intersection. – Dietrich Burde May 14 '20 at 18:11
  • minor note: you shouldn't say "the" basis, for a basis of a vector space is (mostly) not unique. – Aryaman Maithani May 14 '20 at 18:16
  • Please show your efforts on the question to avoid it being closed or heavily downvoted. See here to know how to ask a good question on this website. (Please take note that this website does not do your homework for you) – sai-kartik May 14 '20 at 18:29
  • This question is a duplicate of many others already on this site. There are several examples of how to compute a basis for the intersection of subspaces in the handy list of related questions at right. https://math.stackexchange.com/q/25371/265466 has a good description of the method. – amd May 14 '20 at 18:31
  • @AryamanMaithani - "mostly"? Bases are never unique. – Paul Sinclair May 14 '20 at 23:45
  • @PaulSinclair - that is not true. Consider any zero dimensional vector space. That only has $\emptyset$ as basis. If that doesn't satisfy you, you may consider $\mathbb F_2$ as a vector space over $\mathbb F_2$ and see that it has a unique basis. – Aryaman Maithani May 15 '20 at 06:11
  • @AryamanMaithani - well, that's why I asked. You got me. There is one exception, or two depending on exactly how one defines "basis". – Paul Sinclair May 15 '20 at 12:18

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