Question #2576

B=(x[sup]2[/sup]+x, x[sup]2[/sup]-2, x[sup]2[/sup]+2x-1) is a subset of the vector space P2 of polynomials of degree no larger than two,T(x[sup]2[/sup]+x) =(1,-2) T(x[sup]2[/sup]-2) = (4,1)& & T(x[sup]2[/sup]+2x-1) = (2,-1)what is matrix representation for T with respect to the bases B for P2 and S=(1,0)(0,1) for R[sup]2[/sup]?
With T given as in the above question, calculate T(7x[sup]2[/sup]+3x-2).

Expert's answer

The matrix of T is

<img src="/cgi-bin/mimetex.cgi?%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%201%20&%204%20&%202%5C%5C%20-2%20&%201%20&%20-1%20%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="\left ( \begin{matrix} 1 & 4 & 2\\ -2 & 1 & -1 \end{matrix} \right )">

Notice P2 has the following standard basis S = (x^2, x, 1).

Hence with respect to S the elements of B have the following coordinates:

(1,1,0), (1,0,-2), (1,2,-1).

We have to find the image of the vector V=(7,3,-2) (these coordinates are given in standard basis) under T.

For this we have to express vector V through basis B.

It can be calculated (manually or with some software) that the inverse matrix to

<img src="http://latex.codecogs.com/gif.latex?A%20=%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%201%20&%201%20&%201%5C%5C%201%20&%200%20&%202%5C%5C%200%20&%20-2%20&%20-1%20%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="A =\left ( \begin{matrix} 1 & 1 & 1\\ 1 & 0 & 2\\ 0 & -2 & -1 \end{matrix} \right )">

is

<img src="http://latex.codecogs.com/gif.latex?A%5E%7B-1%7D%20=%5Cfrac%7B1%7D%7B3%7D%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%204%20&%20-1%20&%202%5C%5C%201%20&%20-1%20&%20-1%5C%5C%20-2%20&%202%20&%20-1%20%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="A^{-1} =\frac{1}{3} \left ( \begin{matrix} 4 & -1 & 2\\ 1 & -1 & -1\\ -2 & 2 & -1 \end{matrix} \right )">

Then matrix of T is base S is

<img src="http://latex.codecogs.com/gif.latex?TA%5E%7B-1%7D%20=%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%0A1%20&%204%20&%202%5C%5C%20%0A-2%20&%201%20&%20-1%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%20%29%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%0A4%20&%20-1%20&%202%5C%5C%20%0A1%20&%20-1%20&%20-1%5C%5C%0A-2%20&%202%20&%20-1%20%0A%5Cend%7Bmatrix%7D%20%5Cright%20%29%20=%20%5C%5C%0A=%20%5Cfrac%7B1%7D%7B3%7D%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%0A4%20&%20-1%20&%20-4%5C%5C%20%0A-5%20&%20-1%20&%20-4%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="TA^{-1} =\left ( \begin{matrix}

1 & 4 & 2\\

-2 & 1 & -1\\

\end{matrix} \right ) \times \frac{1}{3} \left ( \begin{matrix}

4 & -1 & 2\\

1 & -1 & -1\\

-2 & 2 & -1

\end{matrix} \right ) = \\

= \frac{1}{3} \left ( \begin{matrix}

4 & -1 & -4\\

-5 & -1 & -4\\

\end{matrix} \right )">

Therefore

lt;img src="http://latex.codecogs.com/gif.latex?T%28V%29%20=%20%5Cfrac%7B1%7D%7B3%7D%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%204%20&%20-1%20&%20-4%5C%5C%20-5%20&%20-1%20&%20-4%5C%5C%20%5Cend%7Bmatrix%7D%20%5Cright%20%29%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%207%20%5C%5C%203%5C%5C%20-2%20%5Cend%7Bmatrix%7D%20%5Cright%20%29%20=%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%2011%20%5C%5C%2010%20%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="T(V) = \frac{1}{3} \left ( \begin{matrix} 4 & -1 & -4\\ -5 & -1 & -4\\ \end{matrix} \right )\left ( \begin{matrix} 7 \\ 3\\ -2 \end{matrix} \right ) = \left ( \begin{matrix} 11 \\ 10 \end{matrix} \right )">

<img src="/cgi-bin/mimetex.cgi?%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%201%20&%204%20&%202%5C%5C%20-2%20&%201%20&%20-1%20%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="\left ( \begin{matrix} 1 & 4 & 2\\ -2 & 1 & -1 \end{matrix} \right )">

Notice P2 has the following standard basis S = (x^2, x, 1).

Hence with respect to S the elements of B have the following coordinates:

(1,1,0), (1,0,-2), (1,2,-1).

We have to find the image of the vector V=(7,3,-2) (these coordinates are given in standard basis) under T.

For this we have to express vector V through basis B.

It can be calculated (manually or with some software) that the inverse matrix to

<img src="http://latex.codecogs.com/gif.latex?A%20=%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%201%20&%201%20&%201%5C%5C%201%20&%200%20&%202%5C%5C%200%20&%20-2%20&%20-1%20%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="A =\left ( \begin{matrix} 1 & 1 & 1\\ 1 & 0 & 2\\ 0 & -2 & -1 \end{matrix} \right )">

is

<img src="http://latex.codecogs.com/gif.latex?A%5E%7B-1%7D%20=%5Cfrac%7B1%7D%7B3%7D%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%204%20&%20-1%20&%202%5C%5C%201%20&%20-1%20&%20-1%5C%5C%20-2%20&%202%20&%20-1%20%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="A^{-1} =\frac{1}{3} \left ( \begin{matrix} 4 & -1 & 2\\ 1 & -1 & -1\\ -2 & 2 & -1 \end{matrix} \right )">

Then matrix of T is base S is

<img src="http://latex.codecogs.com/gif.latex?TA%5E%7B-1%7D%20=%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%0A1%20&%204%20&%202%5C%5C%20%0A-2%20&%201%20&%20-1%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%20%29%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%0A4%20&%20-1%20&%202%5C%5C%20%0A1%20&%20-1%20&%20-1%5C%5C%0A-2%20&%202%20&%20-1%20%0A%5Cend%7Bmatrix%7D%20%5Cright%20%29%20=%20%5C%5C%0A=%20%5Cfrac%7B1%7D%7B3%7D%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%0A4%20&%20-1%20&%20-4%5C%5C%20%0A-5%20&%20-1%20&%20-4%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="TA^{-1} =\left ( \begin{matrix}

1 & 4 & 2\\

-2 & 1 & -1\\

\end{matrix} \right ) \times \frac{1}{3} \left ( \begin{matrix}

4 & -1 & 2\\

1 & -1 & -1\\

-2 & 2 & -1

\end{matrix} \right ) = \\

= \frac{1}{3} \left ( \begin{matrix}

4 & -1 & -4\\

-5 & -1 & -4\\

\end{matrix} \right )">

Therefore

lt;img src="http://latex.codecogs.com/gif.latex?T%28V%29%20=%20%5Cfrac%7B1%7D%7B3%7D%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%204%20&%20-1%20&%20-4%5C%5C%20-5%20&%20-1%20&%20-4%5C%5C%20%5Cend%7Bmatrix%7D%20%5Cright%20%29%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%207%20%5C%5C%203%5C%5C%20-2%20%5Cend%7Bmatrix%7D%20%5Cright%20%29%20=%20%5Cleft%20%28%20%5Cbegin%7Bmatrix%7D%2011%20%5C%5C%2010%20%5Cend%7Bmatrix%7D%20%5Cright%20%29" title="T(V) = \frac{1}{3} \left ( \begin{matrix} 4 & -1 & -4\\ -5 & -1 & -4\\ \end{matrix} \right )\left ( \begin{matrix} 7 \\ 3\\ -2 \end{matrix} \right ) = \left ( \begin{matrix} 11 \\ 10 \end{matrix} \right )">

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