Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is an example from a text (Linear Algebra, Freidberg). I am trying to follow along, and I feel like I should know this from vector calc but I am missing something silly.

The example is:

Define the Linear transformation $T:P_2(R)\rightarrow M_{2x2}(R)$ by

$$T(\,f(x)) = \begin {pmatrix} f(1)-f(2) & 0 \\ 0 & f(0) \end{pmatrix} \DeclareMathOperator{\span}{span}$$

since $\beta = ${$1, x, x^2$} is a basis for $P_2(R)$ we have \begin{align*} R(T) &= \span(T(\beta)) \\ &= \span(\{T(1), T(x), T(x^2)\}) \\ &= \span\left(\left\{\begin {pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin {pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix}, \begin {pmatrix} -3 & 0 \\ 0 & 0 \end{pmatrix} \right\}\right) \\ &= \span\left(\left\{\begin {pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin {pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix} \right\}\right). \end{align*} I wasn't sure how they got from the matrix defined above to the ones below. Neither $f(1)$ nor $f(2)$ are really defined, and plugging an $x$ or $x^2$ in there doesn’t help.

This is frustrating, because I am trying to just get through an example so I can figure out this stuff. The book says “thus we have found a basis for $R(T)$”, but it is far from obvious to me. I understand they are using the theorem that says $$R(T) = \span(T(\beta)) = \span(\{T(v_1), T(v_2)...T(v_n)\}),$$ where the $v_i$ are vectors that form a basis.

It’s getting through the matrix calculation above I am struggling with.

share|improve this question
    
Write \left\{ , \right\} for enclosing things in parentheses of the right size. –  DonAntonio Jun 22 '13 at 20:33
    
I went in and fixed the MathJax for you. Some tips: Use the pmatrix environment instead of matrix, as this adds brackets around the matrix for you. Using \{…\} means you don’t need to drop out of a maths environment to have curly braces. \left\{…\right\} get the correct sizing. The align environment helps get multiple lines all lined up correctly. I used a \DeclareMathOperator to get the spacing and formatting of the “span”s correct. –  alexwlchan Jun 22 '13 at 20:36
    
thanks alex. when one is busy panicking over stuff like this MathJax can be a bit daunting :-) –  Jesse Jun 22 '13 at 20:41

2 Answers 2

up vote 2 down vote accepted

Your transformation is from the space of polynomials of degree at most 2 to the space of $2\times 2$ matrices with real coefficients.

They tell you that the polynomials $1$, $x$, and $x^2$ are a basis for $P_2(\mathbb{R})$; the next three matrices are just the results of applying your transformation to these three polynomials:

If $f(x)=1$ for all $x$, then $$ T(\,f(x))=\begin{pmatrix}f(1)-f(2) & 0\\0 & f(0)\end{pmatrix}=\begin{pmatrix}1-1 & 0\\ 0 & 1\end{pmatrix}=\begin{pmatrix}0&0\\0&1\end{pmatrix}. $$

If $f(x)=x$, then $$ T(\,f(x))=\begin{pmatrix}f(1)-f(2) & 0\\0 & f(0)\end{pmatrix}=\begin{pmatrix}1-2 & 0\\0 & 0\end{pmatrix}=\begin{pmatrix}-1 & 0\\0 & 0\end{pmatrix}.$$

Finally, if $f(x)=x^2$, then $$ T(\,f(x))=\begin{pmatrix}f(1)-f(2) & 0\\0 & f(0)\end{pmatrix}=\begin{pmatrix}1-4 & 0\\0 & 0\end{pmatrix}=\begin{pmatrix}-3 & 0\\0 & 0\end{pmatrix}. $$

share|improve this answer
    
thanks, that helps a ton -- I can't believe I didn't see that right in front of me. –  Jesse Jun 22 '13 at 20:42
    
By the way the reason only the first two matrices are necessary to show it is a basis is because the one with -3 in it is a scalar multiple of the second one, yes? –  Jesse Jun 22 '13 at 20:44
    
Yep - because the one with $-3$ is a scalar multiple of the other one, and therefore contained in its span. –  Nicholas R. Peterson Jun 22 '13 at 20:48

Let's take a look at how the elements of $T(f)$ are defined. We can easily show that $T(\alpha f+g) = \alpha T(f)+T(g)$ for, in fact, any functions $f$, $g$ and $\alpha\in \mathbb C$, because $(\alpha f+g)(x) = \alpha f(x)+g(x)$; this allows to say that $T$ is indeed linear. Apparently, $R(T)=span(T(1),T(x),T(x^2))$, (we look at the basis in $P_2[x]$), so we study what becomes of $T(1)$, $T(x)$, $T(x^2)$. After having obtained $R(T)$ we easily find its basis.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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