Suppose i have a linear mapping from $L:\mathbb{R^2}_v \rightarrow \mathbb{R^2}_w$ and in $ \mathbb{R^2}_v$ i chose the standard basis and a vector is represented by $\begin{bmatrix}4\\3 \end{bmatrix}$ , since it is a standard basis then it's co-ordinates are same but i want to chose in $\mathbb{R^2}_w$ a different basis, for example $(\begin{bmatrix}6\\0 \end{bmatrix}$$\begin{bmatrix}0\\3\end{bmatrix})$.
I can compute the transformation matrix but i don't understand how this basis differs from my stadard basis of $\mathbb{R^2}_v$ ? These just two vectors with length $(6,3)$ instead of $(1,1)$ and what is the point to do this? Any example would be helpful.
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$\begingroup$So the basis is different from the standard basis $ \{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \}$. You are right that the directions do not change but that the lengths of the basis vectors do change. To write the vector with respect to the new basis should "shorten" it. Your vector $ \begin{bmatrix} 4 \\ 3 \end{bmatrix} $ becomes $ \begin{bmatrix} 2/3 \\ 1 \end{bmatrix} $. With this example you can see that the top entry was shortened (divided by $6$) and the bottom entry was shortened (divided by $3$) as well. Transformations are important because they can simplify problems: for this particular example, it is possible to choose a basis for $w$ that reduces the vector $ \begin{bmatrix} 4 \\ 3 \end{bmatrix} $ to a vector as simple as $ \begin{bmatrix} \sqrt{2}/2 \\ \sqrt{2}/2 \end{bmatrix}$.
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