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Let $$T : P_4 \rightarrow P_{3} $$ be given by : $$ T(a_0 + a_1x + a_2x^2 + a_3x^3) = (a_0-a_1+2a_2-a_3) + (-a_0+3a_1 - 2a_2+3a_3)x + (2a_0 - 3a_1+ 5a_2)x^2 + (3a_0 - a_1 + 7a_2 + 2a_3)x^3 $$

Find a basis for R(T).

I am unsure of how to find a matrix of this linear transformation in order to find the basis for R(T).

I am able to solve problems such as this one posted: Finding range of a linear transformation. However, I am confused as to the fact that I am only given the singular equation for the transformation. Is it equivalent to write:

$$ T(a_0 , a_1x , a_2x^2 , a_3x^3) = ((a_0-a_1+2a_2-a_3), (-a_0+3a_1 - 2a_2+3a_3), (2a_0 - 3a_1+ 5a_2), (3a_0 - a_1 + 7a_2 + 2a_3) ) $$

and then find the range similarly to the linked problem?

Thank you!

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    $\begingroup$ Why did you reverse the order of the component functions the second time you wrote it down? $\endgroup$ Oct 5, 2014 at 22:55
  • $\begingroup$ Other than that, I'd say yes, it's equivalent. You can think of the polynomial $a_0+a_1x+a_2x^2+a_3x^3$ and the vector $[a_0,a_1,a_2,a_3]^T$ as essentially the same object. $\endgroup$ Oct 5, 2014 at 22:56
  • $\begingroup$ My apologies! I made up the vectors on the spot so I must have switched them around. I will fix that in the problem statement. Thank you! $\endgroup$
    – 01011010
    Oct 5, 2014 at 23:09

1 Answer 1

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The matrix of the linear map with respect to the basis $\{1,x,x^2,x^3\}$ is

$$\left(\begin{array}{rrrr} 1 & -1 & 2 & -1 \\ -1 & 3 & -2 & 3\\ 2 & -3 & 5 & 0\\ 3 & -1 & 7 & 2\end{array}\right).$$

Note that the first column is $T(1)=T\left(\begin{array}{r} 1 \\ 0\\ 0\\ 0\end{array}\right)=\left(\begin{array}{r} 1 \\ -1\\ 2\\ 3\end{array}\right),$ the second column is $T(x)=T\left(\begin{array}{r} 0 \\ 1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{r} -1 \\ 3\\ -3\\ -1\end{array}\right),$ the third column is $T(x^2)=T\left(\begin{array}{r} 0 \\ 0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{r} 2 \\ -2\\ 5\\ 7\end{array}\right),$ and the fourth column is is $T(x^3)=T\left(\begin{array}{r} 0 \\ 0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{r} -1 \\ 3\\ 0\\ 2\end{array}\right).$

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  • $\begingroup$ Thank you! This doesn't seem so tricky anymore. $\endgroup$
    – 01011010
    Oct 5, 2014 at 23:14

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