# Linear Transformations and Polynomials

• Nov 8th 2009, 08:22 PM
Shapeshift
Linear Transformations and Polynomials
Consider the transformation of ax^2 + bx + c in P2 to absolute value of a in P0. Show that it does not correspond to a linear transformation by showing that there is no matrix that maps (a,b,c) in R^3 to absolute value of a in R.

Any help is appreciated
• Nov 9th 2009, 03:43 AM
HallsofIvy
Quote:

Originally Posted by Shapeshift
Consider the transformation of ax^2 + bx + c in P2 to absolute value of a in P0. Show that it does not correspond to a linear transformation by showing that there is no matrix that maps (a,b,c) in R^3 to absolute value of a in R.

Any help is appreciated

Actually, I would think that the best way to show this is not a linear transformation would be to show that $\displaystyle L(\alpha v)$ is not always equal to $\displaystyle \alpha L(v)$. And do that by taking $\displaystyle \alpha= -1$.

However, since you are asked to do this by showing that there is no matrix representing this linear transformation, take $\displaystyle v_1= 1$, $\displaystyle v_2= x$, and $\displaystyle v_3= x^2$ as basis for P2 and 1 as basis for P0. Then $\displaystyle ax^2+ bx+ c= av_3+ bv_2+ cv_1$. A standard way to construct a matrix for a linear transformation, given a basis for domain and range spaces, is to take that transformation of each of the basis vectors of the domain, in turn, and write the result as a linear combination of the basis vecotrs in the range. The coefficients are the columns of the matrix. Here, $\displaystyle L(x^2)= L(1x^2+ 0x+ 0)= 1$, L(x)= $\displaystyle L(0x^2+ 1x+ 0)= 0$ and $\displaystyle L(1)= L(0x^2+ 0x+ 1)= 0$. That is, if this were a linear transformation, its matrix in the standard bases would be the row matrix [1 0 0].

But $\displaystyle \begin{bmatrix}1 & 0 & 0\end{bmatrix}\begin{bmatrix}-1 \\ 0 \\ 0\end{bmatrix}= -1$ while [tex]L(-x^2)= |-1|= 1.
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