Can the dimension of a basis be less than the space that it spans?

[Pedro1]

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.

 

[I like Serena]

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.

Welcome to MHB, Pedro! :)

Yes, you have a correct basis.

To write a vector in a basis, you need to find a linear combination of your basis vectors.
In your case that is $\mathbf u = 6 \cdot (-\frac{3}{2},1)$.

And to answer your question in the thread title, no, the number of vectors in a basis matches the dimension of the space they span by definition.

 

[Fernando Revilla]

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.

We have $\left[ \begin{matrix} -9 \\ 6 \end{matrix}\right]=6\left[ \begin{matrix} -3/2 \\ 1 \end{matrix}\right]$. This implies that the coordinate vector of $u$ a in terms of $B=\{(-3/2,1)\}$ is $_B=[6]$ (only one coordinate).

P.S. Sorry I Like Serena, I had almost completed my post.

 

[Pedro1]

Thanks for the help.
I kown the the vector $u$ is on the trivial basis of $\mathbb{R^2}$ and so $(-9,6)=-9(1,0)+6(0,1)$. Normaly what I do is write each vector of the trivial basis as a linear combination of the given basis, in this case $B$.
But,

$(1,0)=\lambda\left (-\frac{3}{2},1 \right )$

I couln't find the $\lambda$.

The dimension of $B$ is $1$, but $S \subset \mathbb{R^2}$ have dimension $2$.

 

[Fernando Revilla]

but $S \subset \mathbb{R^2}$ have dimension $2$.

$S$ has dimension 1. :)

 

[Pedro1]

$S$ has dimension 1. :)

Well, I thougt that the dimension was the number of vectors that the trivial basis have. In this case $\mathbb{R^2}$ have $2$.

 

[I like Serena]

Well, I thougt that the dimension was the number of vectors that the trivial basis have. In this case $\mathbb{R^2}$ have $2$.

Yes. $\mathbb{R^2}$ has a trivial basis with 2 basis vectors, implying it has 2 dimensions.
Your S has a basis of 1 vector, implying it has 1 dimension.