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I want to be clear that I am understanding how to construct a matrix $A$ corresponding to specified linear transformation $T$.

I am working through "Introduction to Linear Algebra by Strang" 4th edition.

Here is what is mentioned:

Suppose $T$ transforms the space $V$ ($n$-dimensional) to the space $W$ ($m$-dimensional). We choose a basis for $v_1,\ldots,v_n$ for $V$ and a basis $w_1,\ldots, w_m$ for $W$.

To find the first column of $A$, apply $T$ to the first basis vector $v_1$.

Then $T(v_1) = a_{11}w_1 + \cdots + a_{m1}w_m$ and these numbers $a_{11},\ldots,a_{m1}$ go into the first column of $A$. Transforming $v_1$ to $T(v_1)$ matches multiplying $(1,0,\ldots,0)$ by $A$.

My questions/thoughts

This section opens up by saying this approach works for constructing a matrix for any linear transformation. However, it seems the basis in $n$-dimensional space is stated as being generic (i.e. $v_1,\ldots,v_n$) but if $T(v_1)$ matches multiplying by $(1,0,\ldots,0)$ am I supposed to assume $v_1,\ldots,v_n$ are a standard orthonormal basis in $n$-dimensional space?

I guess I don't see how this example generalizes, but perhaps I am missing something.

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Consider that the components of $v_1$ are determined by $$v_1=1 v_1+0 v_2+\cdots+0 v_n.$$

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  • $\begingroup$ since the $v_i$ are linearly independent $\endgroup$ – janmarqz Apr 3 '18 at 21:24
  • $\begingroup$ But suppose I choose a basis vector in $3$-space to be $v_1$ like $(17,-5,0)$. Clearly, $Av_1$ will be $a_117 + a_2(-5)$ (where $a_i$ are columns of $A$) and involves more than just the first column of $A$. Hence $v_1 = (1,0,0)$ for this example to work, correct? $\endgroup$ – ClownInTheMoon Apr 3 '18 at 21:30
  • $\begingroup$ If $U=17v_1-5v_2$ then you only have calculate $T(U)=17T(v_1)-5T(v_2)$ $\endgroup$ – janmarqz Apr 3 '18 at 21:37
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    $\begingroup$ @ClownInTheMoon Yes. That’s a good summary. As well, the “length” of a unit step might vary from direction to direction. $\endgroup$ – amd Apr 4 '18 at 0:45
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    $\begingroup$ @ClownInTheMoon It's also important to keep in mind that vectors aren’t necessarily tuples of numbers. They could be polynomials or current flows in an electrical network, for instance. Their coordinates, on the other hand, can be assembled into tuples, creating an isomorphism between the vector space and the space of tuples over the scalar field. $\endgroup$ – amd Apr 4 '18 at 0:53
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I always explain it more simply:

You have a linear transformation $ L: \mathbb{R}^n \rightarrow \mathbb{R}^m $. You know that for any vector $ x = \left( \begin{array}{c} \chi_0 \\ \chi_1 \\ \vdots \\ \chi_{n-1} \end{array} \right) = \chi_0 \left( \begin{array}{c} 1\\ 0 \\ \vdots \\ 0 \end{array} \right) + \chi_1 \left( \begin{array}{c} 0\\ 1 \\ \vdots \\ 0 \end{array} \right) = \chi_0 e_0 + \chi_1 e_1 + \cdots $ $$ L( x ) = L( \chi_0 e_0 + \chi_1 e_1 + \cdots ) = \chi_0 L( e_0 ) + \chi_1 L( e_1 ) + \cdots $$ Thus, the linear transformation is completely defined by how the standard basis vectors $ e_j $ are transformed. A matrix is merely a convenient way of then writing this information: the $ j$th column of the matrix equals the vector that results from evaluating $ L( e_j ) $.

What Strang is trying to do here is to generalize this to viewing vectors in different bases.

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