# Math Help - Basis for Subspace

1. ## Basis for Subspace

My homework question is asking me to define the basis of the subspace. S consists of all vectors in the form (x,y,2x,2y) in R4, and the dimension of the subspace.

My book does not do a good job of explaining this so I'm a little lost.

My first guess would be (1, 2, 2, 6) (0, 1 ,0 , 3) and (1, 0, 2, 0). I'm not exactly sure what to do with this or if I'm even on the right track.

2. If all the vectors have the form (x, y, 2x, 2y), then you can immediately see that the dimension of the space is 2. That's because once you have chosen values for x and y, the whole thing is then determined. Dimension 2 means your basis will have two vectors. What would a basis for vectors like (x, y) look like?

3. Originally Posted by nanvinnie
My homework question is asking me to define the basis of the subspace. S consists of all vectors in the form (x,y,2x,2y) in R4, and the dimension of the subspace.

My book does not do a good job of explaining this so I'm a little lost.

My first guess would be (1, 2, 2, 6) (0, 1 ,0 , 3) and (1, 0, 2, 0). I'm not exactly sure what to do with this or if I'm even on the right track.

Assuming you know linear transformations,

Observe that $\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}1 & 0 & 2 & 0 \\0 & 1 & 0 & 2\end{pmatrix} = \begin{pmatrix}x & y & 2x & 2y\end{pmatrix}$

$T \equiv \begin{pmatrix}1 & 0 & 2 & 0 \\0 & 1 & 0 & 2\end{pmatrix}$ is the matrix of the linear transformation(it is not in standard form though).

Prove that the rank of this matrix is 2. Then dim(image(T)) is 2. But image(T) is precisely the set of vectors of the form $\begin{pmatrix}x & y & 2x & 2y\end{pmatrix}$

Once you know the dimension is 2 and the theorem that basis vectors map to basis vectors. we see that (1,0,2,0) and (0,1,0,2) are the basis elements.

Without knowing linear transformations, you could have observed that given any arbitrary vector $\begin{pmatrix}x & y & 2x & 2y\end{pmatrix}$, we can write it as $x\begin{pmatrix}1 & 0 & 2 & 0\end{pmatrix} + y\begin{pmatrix}0 & 1 & 0 & 2\end{pmatrix}$. This means the set $\{\begin{pmatrix}1 & 0 & 2 & 0\end{pmatrix},\begin{pmatrix}0 & 1 & 0 & 2\end{pmatrix}\}$ is a spanning set.

And then also prove that the set $\{\begin{pmatrix}1 & 0 & 2 & 0\end{pmatrix},\begin{pmatrix}0 & 1 & 0 & 2\end{pmatrix}\}$ is linearly independent and that you cant drop either vector from the set.

Thus we have proved that the set is a minimal spanning set and hence a basis. Now since the basis has 2 elements, the dimension of the subspace is 2.