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Does there exist a matrix representation for the linear transformation
$T(x P(x)) = (x-1)P(x)$, where $P(x)$ is the second degree polynomial?

Here, $xP(x)$ are all third degree polynomials that satisfy $P(0) = 0$ and $(x-1) P(x)$ are all third degree polynomials, that satisfy $P(1) = 0$.

I know that a basis for $xP(x)$ is $\{x, x^2, x^3\}$ and basis for $(x-1) P(x)$ is $\{x-1, x^2-x, x^3 - x^2\}$, but can't figure out how to write a matrix for linear transformation.

sergeyz
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2 Answers2

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For the two bases you gave, the matrix corresponding to that transformation is just the identity matrix (since $x$ maps to $x-1$, $x^2$ maps to $x^2-x$, and $x^3$ maps to $x^3-x^2$).

jschnei
  • 1,641
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If we represent $P(x) = ax^2 + bx + c$ as

$$P = \left(\matrix{ [x^3]_P \\ [x^2]_P \\ [x^1]_P \\ [ x^0 ]_P }\right) = \left(\matrix{ 0 \\ a \\ b \\ c }\right)$$

then

$$T = \left(\matrix{1& 0& 0 & 0\\-1 & 1 & 0 & 0\\0 & -1 & 1 & 0\\0 & 0 & -1 & 1}\right)$$

is a matrix representation of the transform since

$$T[xP] = \left(\matrix{1& 0& 0 & 0\\-1 & 1 & 0 & 0\\0 & -1 & 1 & 0\\0 & 0 & -1 & 1}\right)\left(\matrix{ a \\ b \\ c \\ 0 }\right) = \left(\matrix{ a \\ b-a \\ c-b \\ -c }\right)$$

i.e. $T[xP] = ax^3 + (b-a)x^2 + (c-b)x - c = (x-1)P(x)$.

Winther
  • 24,478
  • Beautiful!....Thank you!....This is exactly what I need!....How were you able to come up with the matrix? – sergeyz Dec 12 '14 at 23:54