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Let $T(a_{0}+a_{1}x+a_{2}x^2)=2a_{0}+a_{2}+(a_{0}+a_{1}+a_{2})x+3a_{2}x^2$ be a linear transformation. I need to find the eigen-vectors eigenvalues of $T$.

So, I'm trying to find $[T]_{E}^{E}$ when the base is $E=\{1,x,x^2\}$.

I don't understand how I should use this transformation to do that.

Thanks.

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    in $R^n$, which matrix takes $(a_0, a_1, a_2)$ to $(2a_0 + a_2, a_0 + a_1 + a_2, 3a_2)$? It will be the same matrix as $[T]_E^E$...2011-07-19

3 Answers 3

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You can identify an arbitrary polynomial $a_0+a_1x+a_2x^2$ with the vector $[a_0\ a_1\ a_2]^T$. So $T$ is the mapping $ \begin{bmatrix} a_0 \\ a_1 \\ a_2 \end{bmatrix} \mapsto \begin{bmatrix} 2a_0+a_2 \\ a_0+a_1+a_2 \\ 3a_2\end{bmatrix}. $ The matrix that does just that is $ \begin{bmatrix} 2 & 0 & 1\\ 1 & 1 & 1\\ 0 & 0 & 3\\ \end{bmatrix}. $

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Hint: Write the vector $a_{0}+a_{1}x+a_{2}x^{2}$ in the basis E, and represent it as $\left[\begin{array}{c} a_{0}\\ a_{1}\\ a_{2}\end{array}\right].$

Then notice $\left[\begin{array}{ccc} 2 & 0 & 1\\ 1 & 1 & 1\\ 0 & 0 & 3\end{array}\right]\left[\begin{array}{c} a_{0}\\ a_{1}\\ a_{2}\end{array}\right]=\left[\begin{array}{c} 2a_{1}+a_{2}\\ a_{1}+a_{2}+a_{3}\\ 3a_{2}\end{array}\right].$

What can you conclude from this?

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The columns of the matrix you seek are the coordinates of the images under $T$ of the elements of the basis. So you need only compute $T(1)$, $T(x)$, and $T(x^2)$.

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    $1 = a_0 + a_1x + a_2x^2$ where $a_0 = 1$ and $a_1 = a_2 = 0$. Plug into your formula...2011-07-19