1
$\begingroup$

I'm having a bit of trouble following the logic my professor used to construct the matrix of a given transformation $T$ in class, and was wondering if anyone could share any further intuition or insight.

Given $P_2$, the set of polynomials of degree $\leq2$, define a linear transformation $T:P_2\rightarrow \mathbb{R}^3$ such that $T\big(p(x)\big) = \begin{pmatrix} p(0)\\p(1)\\p(2)\end{pmatrix}$.

The matrix of $T$ with respect to the basis $\mathcal{B}=\left\{1,x,x^2\right\}\\$

is given by $T(1)=\begin{pmatrix} 1\\1\\1\end{pmatrix}$, $T(x)=\begin{pmatrix} 0\\1\\2\end{pmatrix}$, and $T(x^2)=\begin{pmatrix} 0\\1\\4\end{pmatrix}$,

i.e. $\,A=\begin{pmatrix}1&0&0\\1&1&1\\1&2&4\end{pmatrix}$.

I get the whole business of the coefficients on this particular transformation being the "trivial relation" such that $c_1=c_m=0$, and that $ImT=\mathbb{R}^{3}$ such that $T$ is an isomorphism.

What I'm struggling with, quite frankly, is how he got the coefficients for $A$. The coefficients for $p(x)=1$ in particular are a bit counter-intuitive. Is the idea that, if we define $p(x)=1$, then no matter what, $p(0)=p(1)=p(2)=1$?


The notation for expressing the various matrices associated with an isomorphism is also a bit nebulous for me. My textbook defines a $\mathcal{B}$-coordinate transformation as

$T^{-1}\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}=c_1f_1+\cdots + c_nf_n$.

What would the inverse of the transformation $T\big(p(x)\big) = \begin{pmatrix} p(0)\\p(1)\\p(2)\end{pmatrix}$ be?

What would the matrix associated with the inverse be? Or would it be a set of three different polynomials (linear equations)?

For those wondering, we're using Bretscher's Linear Algebra with Applications, 5th edition. Thanks!

$\endgroup$
  • $\begingroup$ $A = [[T(1)]_B|[T(x)]_B|[T(x^2)]_B]$ where $B=\{ e_1,e_2,e_3 \}$ is just the standard basis for $\mathbb{R}^3$. More generally, you just find the coordinate vectors of the image of each vector in the domain basis and glue them together. There are good reasons for doing this, but you ought to have seen them in lecture I think... $\endgroup$ – James S. Cook Mar 22 '15 at 4:42
  • $\begingroup$ My silly notation $[v]_B = v$ in this case since the coordinate system in the codomain is just the plain-old-standard Cartesian coordinate system. $\endgroup$ – James S. Cook Mar 22 '15 at 4:43
1
$\begingroup$

To give you a much more complicated example. Let us consider $T(f(x)) = f'(x)$ where we consider $T:P_2 \rightarrow P_2$ and to be perverse let's use $\beta = \{ 1,x,x^2 \}$ as the domain basis, but $\gamma = \{ x^2,x,1 \}$ as the codomain basis. To understand $T$ I like to consider $f(x)=a+bx+cx^2$ and see what happens: $$ T(f(x)) = b+2cx $$ Then the coordinate vector of the image is easy enough to see: $$ [T(f(x))]_{\gamma}= [b+2cx]_{\gamma} = [0(x^2)+2c(x)+b(1)]_{\gamma} = [0,2c,b]^T. $$ Then, the matrix $[T]_{\beta,\gamma}$ is the matrix which when multiplied on $$[f(x)]_{\beta} = [a+bx+cx^2]_{\beta} = [a,b,c]^T$$ yields $[T(f(x))]_{\gamma}=[0,2c,b]^T$. A moments reflection yields: $$ [T]_{\beta, \gamma} = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 1 & 0\end{array}\right]. $$ Alternatively, just use: $$ [T]_{\beta, \gamma} = [ [T(1)]_{\gamma} | [T(x)]_{\gamma}| [T(x^2)]_{\gamma} ] = [[0]_{\gamma}|[1]_{\gamma}|[2x]_{\gamma}] = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 1 & 0\end{array}\right].$$ Some books also use $[T]_{\beta}^{\gamma}$ to reflect the differing roles the domain and codomain bases play especially in regard to coordinate change. Perhaps your text does that.

$\endgroup$
  • $\begingroup$ The text doesn't explicitly use the terms "domain" and "codomain", but I think I understand. Follow-up question: if we define some other basis $\mathcal{C}=\left\{x^2,xy,y^2\right\}$ and $T(x)=\begin{pmatrix}f(1,0)\\f(0,1)\\f(1,1)\end{pmatrix}$, does the computation still move along similar lines? see here for more. $\endgroup$ – Benjamin Loya Mar 22 '15 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.