$$W_1=\operatorname{span}\left(\begin{pmatrix}1&1\\ 0&0\end{pmatrix},\begin{pmatrix}3&1\\ -1&0\end{pmatrix}\right)$$

$$W_2=\operatorname{span}\left(\begin{pmatrix}1&1\\ 1&0\end{pmatrix},\begin{pmatrix}2&-1\\ -1&1\end{pmatrix}\right)$$

For the given spanning sets find, the dimension of $:W_1\cap W_2$ and it's base. I took some random scalars from the respective field and rewrote the spanning sets and set the equal and got a system of equations is that them correct thing to do?

$$\exists \alpha ,\beta ,\gamma ,\delta \in \mathbb{F}$$

$\begin{pmatrix}\alpha +3\beta &\alpha +\beta \\ -\beta &0\end{pmatrix}=\begin{pmatrix}\gamma +2\delta &\gamma -\delta \\ \gamma +\delta &\delta \end{pmatrix}$

$$\alpha \:+3\beta \:=\gamma \:+2\delta \:,\:\alpha \:+\beta \:=\gamma \:-\delta \:,\:-\beta \:=\gamma \:+\delta \:,\:0=\delta \:$$

  • $\begingroup$ In general $\dim W_1\cap W_2=\dim W_1+\dim W_2-\dim(W_1+W_2)$. $\endgroup$ – Spenser Jul 1 '16 at 12:49
  • $\begingroup$ WhAT would be W1+W2? $\endgroup$ – gvidoje Jul 1 '16 at 12:50
  • $\begingroup$ A set of vectors is linearly dependent if and only if one of the elements is a linear combination of the others. In your case, you can check case by case that none of the four matrices is a linear combination of the three others. Hence, they are linearly independent, so $\dim(W_1+W_2)=4$. $\endgroup$ – Spenser Jul 1 '16 at 12:54
  • $\begingroup$ Okay, but would solving the equation that i got be wrong? $\endgroup$ – gvidoje Jul 1 '16 at 12:56
  • 1
    $\begingroup$ Yes that is also a good approach. You should find $\alpha=\beta=\gamma=\delta=0$, which means $\dim W_1\cap W_2=0$. Both approach thus yield to the same solution. $\endgroup$ – Spenser Jul 1 '16 at 12:59

By your fourth equation, we have $\delta=0$, so by your first equation, we have $\alpha+\beta=\gamma$ and by your second equation, we have $\alpha+3\beta=\gamma$. Subtract the second equation by the first equation to get $2\beta=0 \implies \beta=0$. Now, by the third equation, we have $0=\gamma$ and, going back to the first equation, we get $\alpha=0$.

Thus, $\alpha=\beta=\delta=\gamma=0$, so the only element in $W_1 \cap W_2$ is $\mathbf 0$, giving us $\dim(W_1 \cap W_2)=0$.

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