Consider the linear transformation $\phi : \mathbb{R}^2 \to \mathbb{R}^2$ defined by $\phi (x,y) = (x+y, x- 2y)$. Let $p(x) = x^2 -2x + 1$. Does $p(\phi)$ make sense and if yes what is it?
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$\begingroup$ $\phi^2=\phi\circ \phi$ etc. $\endgroup$– JP McCarthyFeb 13, 2013 at 1:09
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$\begingroup$ If $p$ is a polynomial then you cannot evaluate a polynomial at a vector in $\mathbb R^2$. But a linear transformation "is" a matrix, and $p(\phi)$ may denote $p(\textrm{this matrix})$, which is a matrix. $\endgroup$– BreninFeb 13, 2013 at 1:12
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2 Answers
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Yes, one can make sense of this according to the following: if $\phi: X \rightarrow X$ is a function on a vector space (not even necessarily linear), and if $p(x) = ax^2 + bx + c$ is a polynomial, then $$ p(\phi) = a (\phi \circ \phi) + b \phi + c \, \mathrm{Id}_X $$ where $\phi \circ \phi$ indicates composition of $\phi$ with itself, and $\mathrm{Id}_X$ is the identity map. We can generalize this appropriately to any polynomial $p$.
In your specific case, you can compute this in a variety of ways, either by hand, or by first finding a matrix representation of $\phi$, call it $A$, and then $p(\phi)$ has a matrix representation in the same basis given by $A^2 - 2A + I$, where $I$ is the identity matrix of the appropriate size.
answered Feb 13, 2013 at 1:09
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$\begingroup$ Fair enough! Thank you very much! $\endgroup$– user44069Feb 13, 2013 at 1:12
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If I have understood your problem, yes, it makes sense. Consider that linear transformations over a vector field form an algebra over a field. You can see easily changing point of view and considering the matrix representation of your transformation. If $A$ is the matrix that represents $\phi$ in the canonical basis of $\mathbb{R}^{2}$ then $p(\phi)$ is represented by $AA+ 2A+ \mathbb{1}$
