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$B=\{e_1,e_2,e_3\}=\{(1,0,0),(0,1,0),(0,0,1)\}$

To find matrices in $span(e_1+e_2,e_3)\cap span(e_1,e_2+e_3)$ we solve a system: $a \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}=b\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ \end{bmatrix}\Rightarrow a=b=0\Rightarrow$ $span(e_1+e_2,e_3)\cap span(e_1,e_2+e_3)=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\Rightarrow$

$\dim (span(e_1+e_2,e_3)\cap span(e_1,e_2+e_3))=0$.

Question: What is a basis of $span(e_1+e_2,e_3)\cap span(e_1,e_2+e_3)$?

Is it $\left\{ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix}\right\}$ or $\left\{ \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}\right\}$?

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    $\begingroup$ The zero vector cannot be part of any basis. The empty set $\;\emptyset\;$ is the (only) basis for the zero subspace. And you calculation is wrong: at least the vector $\;e_1+e_2+e_3\;$ is in both spans. $\endgroup$
    – DonAntonio
    Mar 7, 2016 at 10:10
  • $\begingroup$ @Joanpemo Could you show the method for finding $\cap$ of two spans? $\endgroup$
    – user300048
    Mar 7, 2016 at 10:14

1 Answer 1

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Let $\;a,b,\alpha,\beta\in\Bbb R\;$ , so some vector belongs to that intersection iff

$$a(e_1+e_2)+be_3=\alpha e_1+\beta(e_2+e_3)\iff (a-\alpha)e_1+(a-\beta)e_2+(b-\beta)e_3=0\iff$$

$$\iff a=b=\alpha=\beta$$

and thus

$$\text{Span}\{e_1+e_2,e_3\}\cap\text{Span}\{e_1,e_2+e_3\}=\text{Span}\left\{\;\begin{pmatrix}1\\1\\1\end{pmatrix}\;\right\}$$

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  • $\begingroup$ I don't understand how did you get $span(1,1,1)$ when $a=b=\alpha=\beta$ should be equal to zero? $\endgroup$
    – user300048
    Mar 7, 2016 at 10:27
  • $\begingroup$ @user300044 Look at the left side of the first equation: it has to be $\;a=b\;$ , so any vector in the intersection is of the form $$\;a(e_1+e_2+e_3)=a\begin{pmatrix}1\\1\\1\end{pmatrix}$$ Why do you think $\;a=b=\beta=\alpha\;$ should be zero? $\endgroup$
    – DonAntonio
    Mar 7, 2016 at 10:28
  • $\begingroup$ Is it correct that to form a basis, we need to find two more vectors that are linearly independent and span $\cap$ of those two spans? The dimension of $\cap$ is then $3$. $\endgroup$
    – user300048
    Mar 7, 2016 at 10:57
  • $\begingroup$ @user300044 I'm not sure I understand your question. The dimension of the intersection cannot be three else it'd be the whole space, which is absurd since each ofthose spans is generated by ONLY two vectors so the dimension is at most two. What is shown in my answer is that the dimension is actually one. $\endgroup$
    – DonAntonio
    Mar 7, 2016 at 11:04
  • $\begingroup$ And a basis is $(1,1,1)$, right? $\endgroup$
    – user300048
    Mar 7, 2016 at 11:11

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