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taoy1

What is a linear map?

  • Previous slide: maps vectors to vectors and preserves weighted sums.

Or

  • This slide: for a map $f(\mathbf{u})$ between $R^m$ to $R^n$, $f(u_{1}, ..., u_{m})$ can be expressed as $\sum_{n=1}^{m} u_i\mathbf{a_i}$, while $a_{i}, i \in 1..m $ are vectors in $R^n$,.

These two conditions are equivalent.

$$ f(\mathbf{u}) = f(u_1\mathbf{e_1}+...+u_m\mathbf{e_m}) $$ $$ = f(u_1\mathbf{e_1})+...+f(u_m\mathbf{e_m}) $$ $$ = u_1 f(\mathbf{e_1}) + ... + u_m f(\mathbf{e_m}) $$ let $a_i$ be $f(\mathbf{e_i}), i \in 1..m $ $$ f(\mathbf{u}) = \sum_{n=1}^{m} u_i\mathbf{a_i} $$

Zihan

For the specific example at the bottom, where $f(x): \mathbb{R}^{2} \to \mathbb{R}^{3}$, can we consider ${a}_1$ and ${a}_2 $ are three-dimentional? If $a = ({a}_1,{a}_2,{a}_3,{a}_4)$, will the map be $f(x): \mathbb{R}^{2} \to \mathbb{R}^{4}$? Is this where 3 comes from, the dimention of a?

keenan

@Zihan: Yes, if you express a linear map as a linear combination $\sum_{i=1}^m u_i \mathbf{a}_i$ of m different n-dimensional vectors $\mathbf{a}_1, \ldots, \mathbf{a}_m \in \mathbb{R}^n$, then indeed you have a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$: for each collection of coordinates $(u_1,\ldots,u_m)$ you get a point in $\mathbb{R}^n$. Likewise, *any* map from $\mathbb{R}^m$ to $\mathbb{R}^n$ can be expressed as a linear combination of vectors; if you like thinking about matrices, these are the columns of the matrix representing the linear map.