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Let S be a subspace of $\mathbb{R^2}$, such that $S=\{(x,y):2x+3y=0 \}$.

Find a basis,$B$, for $S$ and write $u=(-9,6)$ in the $B$ basis.

So, I started to solve $2x+3y=0$ for $x$ and I got $x=-\frac{3}{2}y$. Then I could write,

$\left[ \begin{matrix} x \\ y \end{matrix}\right] = \left[ \begin{matrix} -\frac{3}{2} y \\ y \end{matrix} \right] = y\left[ \begin{matrix} -\frac{3}{2} \\ 1 \end{matrix} \right]$

So I was force to conclued that S is spaned by the vector $(-\frac{3}{2},1)$. My doubt is if it is correct to assume that $\{(-\frac{3}{2},1) \}$ is a basis of $S$. Because, in this circunstances, I don't know how to write the vector $u$ in the basis $B$.

Thanks for the help.

Find a basis,$B$, for $S$ and write $u=(-9,6)$ in the $B$ basis.

So, I started to solve $2x+3y=0$ for $x$ and I got $x=-\frac{3}{2}y$. Then I could write,

$\left[ \begin{matrix} x \\ y \end{matrix}\right] = \left[ \begin{matrix} -\frac{3}{2} y \\ y \end{matrix} \right] = y\left[ \begin{matrix} -\frac{3}{2} \\ 1 \end{matrix} \right]$

So I was force to conclued that S is spaned by the vector $(-\frac{3}{2},1)$. My doubt is if it is correct to assume that $\{(-\frac{3}{2},1) \}$ is a basis of $S$. Because, in this circunstances, I don't know how to write the vector $u$ in the basis $B$.

Thanks for the help.

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