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The question I'm answering is as follows:

Let $ T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a linear transformation such that $ T(1,1) = (2,1) $ and $ T(0,2) = (2,8) $. Find a formula for $ T(a,b) $ where $ (a,b) \in \mathbb{R}^2 $.

Earlier we proved that $\{(1,1), (0,2)\}$ spans $\mathbb{R}^2$. I used this when trying to find a formula for $T$. My working is:

$T(a(1,1) + b(0,2)) = aT(1,1) + bT(0,2) $

Because $T$ is linear. Thus:

$ T(a(1,1) + b(0,2)) = a(2,1) + b(2,8) = T(a,b)$

Is this correct? It seems a bit too easy and so I'm wondering if I missed anything.

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er I think you found $T(a, 2b + a)$, but otherwise yeah –  uncookedfalcon Aug 16 '12 at 4:38
    
oh duh! Thanks for that –  user1520427 Aug 16 '12 at 4:43

3 Answers 3

up vote 3 down vote accepted

You’ve gone astray at the very end: $a(1,1)+b(0,2)=(a,a)+(0,2b)=(a,a+2b)$, so what you’ve calculated is $T(a,a+2b)$, not $T(a,b)$.

You know that $T(0,2)=(2,8)$, and $(0,1)=\frac12(0,2)$, so $T(0,1)=\frac12T(0,2)=\frac12(2,8)=(1,4)$. You also know that $(1,0)=(1,1)-(0,1)$, so $T(1,0)=T(1,1)-T(0,1)=(2,1)-(1,4)$. Finally, you know that $(a,b)=a(1,0)+b(0,1)$, so $T(a,b)=aT(1,0)+bT(0,1)$. Can you finish it from here?

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We have $$(1,0)=(1,1)-\frac{1}{2}(0,2)\qquad\text{and} \qquad(0,1)=\frac{1}{2}(0,2).\tag{$1$}$$

Note that $(a,b)=a(1,0)+b(0,1)$. So $$T(a,b)=aT(1,0)+bT(0,1).$$

Now use the values of $T(1,1)$ and $T(0,2)$, and Equations $(1)$, to find $T(1,0)$ and $T(0,1)$, and simplify a bit.

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The matrix representation of $T$ with respect to the bases $\{(1,1),(0,2)\}$ (for the domain) and $\{(1,0),(0,1)\}$ (for the co-domain) is $$\left( \begin{array}{cc} 2 & 2 \\ 1 & 8 \end{array} \right).$$ Using the fact that $a(1,1) + b(0,2) = (a,a+2b)$, the transition matrix from the basis $\{(1,1),(0,2)\}$ to the standard basis $\{(1,0),(0,1)\}$ is $$\left( \begin{array}{cc} 1 & 0 \\ 1 & 2 \end{array} \right).$$ The matrix representation of $T$ with respect to the standard basis (for both the domain and co-domain) is then $$\left( \begin{array}{cc} 2 & 2 \\ 1 & 8 \end{array} \right)\left( \begin{array}{cc} 1 & 0 \\ 1 & 2 \end{array} \right)^{-1} = \frac{1}{2}\left( \begin{array}{cc} 2 & 2 \\ 1 & 8 \end{array} \right)\left( \begin{array}{cc} 2 & 0 \\ -1 & 1 \end{array} \right) = \left( \begin{array}{cc} 1 & 1 \\ -3 & 4 \end{array} \right).$$ Thus $T(a,b) = (a+b,-3a+4b).$

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