0
$\begingroup$

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

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

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?

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

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

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $\dim(W_1 \cap W_2)=0$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.