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I'm trying to solve a linear transformation problem.

Let $ \psi: \mathbb R_3 [x] \to \mathbb R_4 [x] $ be defined by $ \psi : p(x) \mapsto x^4 p(1/x)+p(x)$

Q) Show that $\psi$ is a linear transformation and find the bases for ker($\psi$) and Im($\psi$).

I know that to show it's a linear transformation you must show it satisfies the conditions: $\phi(x+y)=\phi(x)+\phi(y) $ and $ \phi(\lambda x)= \lambda \phi(x)$ but don't know how to go about solving it.

Your help will be greatly appreciated...

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  • $\begingroup$ Why don't you try to evaluate $\psi( p(x) + q(x))$ and $\psi (\lambda p(x))$ for polynomials $p(x), q(x)$ in $\mathbb R_3 [x]$? (Note that you can actually combine the two conditions needed to show linearity into the condition $\psi( \lambda x + \mu y) = \lambda \psi(x) + \mu \psi(y) $). $\endgroup$
    – Daniel Freedman
    Apr 25, 2012 at 0:55
  • $\begingroup$ What's $\mathbb R_3 [x]$? Polynomials over the reals of degree at most 3? $\endgroup$
    – lhf
    Apr 25, 2012 at 1:06
  • $\begingroup$ @DanielFreedman, or just $\psi( \lambda x + y) = \lambda \psi(x) + \psi(y)$. $\endgroup$
    – lhf
    Apr 25, 2012 at 1:06
  • $\begingroup$ I'm not sure you understand what $\psi$ does. Pick a polynomial of degree at most 3, for example, $p(x)=2x^3+3x^2+5x+7$. Are you able to compute $\psi(p)$? $\endgroup$
    – Gerry Myerson
    Apr 25, 2012 at 2:08
  • $\begingroup$ @DanielFreedman how do i got about showing linearity? $\endgroup$
    – Ricky Rozay
    Apr 25, 2012 at 13:39

1 Answer 1

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Let's think about what $\psi$ does to a few polynomials.

Say $p(x) = x^3 -2x$. Then $p(1/x) = \frac{1}{x^3} - 2\frac{1}{x}$, so $$x^4p(1/x) = x^4\left(\frac{1}{x^3}-2\frac{1}{x}\right) = x - 2x^3.$$ In particular, note that it is a polynomial, even though the computation involved something which is not a polynomial. Since $p(x)$ has degree at most $3$, then the highest power of $x$ that can occur in the denominator of $p(1/x)$ is $x^3$, so multiplying by $x^4$ will guarantee all powers of $x$ that occur are nonnegative. You should also try to make sure that the image always has degree at most $4$ (so that it actually is in $\mathbb{R}_4[x]$).

So, if $p(x) = x^3-2x$, for example, then $$\psi(x) = x^4p(1/x) + p(x) = (x-2x^3) + (x^3-2x) = -x^3 -x.$$

If $p(x)$ and $q(x)$ are polynomials, then remember that $(p+q)(u) = p(u)+q(u)$. So $(p+q)(1/x) = p(1/x) + q(1/x)$, whatever $p(1/x)$ and $q(1/x)$ might be. So, what is $\psi(p+q)$? It is $$\psi(p+q) = x^4\Bigl( (p+q)(1/x)\Bigr) + (p+q)(x) = x^4\Bigl(p(1/x)+q(1/x)\Bigr) + p(x)+q(x).$$

Is it equal to $\psi(p) + \psi(q)$? Well, evaluate and check.

If $p(x)$ is a polynomial and $\alpha$ is a scalar, then $\alpha p$ is the polynomial which, when evaluated at $u$, gives $\alpha p(u)$. So $$\psi(\alpha p) = x^4\Bigl(( \alpha p)(1/x)\Bigr) + (\alpha p)(x) = x^4\alpha p(1/x) + \alpha p(x).$$ What is $\alpha\psi(p)$? Is it equal to $\psi(\alpha p)$?

Once you know that it is a linear transformation (or before, if you want to try to get a better handle on it) see if you can figure out a formula for $\psi(a_0 + a_1x + a_2x^2+a_3x^3)$ in terms of $a_0$, $a_1$, $a_2$, and $a_3$ (equivalently, if you know it is linear, figure out $\psi(1)$, $\psi(x)$, $\psi(x^2)$, and $\psi(x^3)$; why will that be enough?) This should help you find out what the image is, and so a basis for it; and what the kernel is (and so a basis for it).

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  • $\begingroup$ how do you go from finding out ψ(1), ψ(x), ψ(x^2), and ψ(x^3) to finding the Image? $\endgroup$
    – Ricky Rozay
    Apr 25, 2012 at 13:57
  • $\begingroup$ @TundeBaba: Since $1$, $x$, $x^2$, and $x^3$ span the domain, their images span the image. $\endgroup$
    – Arturo Magidin
    Apr 25, 2012 at 15:29
  • $\begingroup$ @ Arturo Magidin what your saying makes sense, but i'm really struggling to understand your thought process and compose the maths behind what your saying. Sorry this is not my strong point. $\endgroup$
    – Ricky Rozay
    Apr 26, 2012 at 0:09
  • $\begingroup$ @TundeBaba: My "tought process" is quite simply to follow the definitions. To check if $\psi$ is actually a funtion, we need to take an element of the domain (in this case, a polynomial of degree at most $3$), and verify that the formula gives an element of the codomain (in this case, a polynomial of degree at most $4$); to check $\psi$ is linear, we need to take two "vectors" $p$ and $q$ (which in this case are polynomials of degree at most $3$) and make sure that $\psi(p+q)$ is the same thing as $\psi(p)+\psi(q)$. Take a vector (a polynomial) $p$, and a scalar $\alpha$ (cont) $\endgroup$
    – Arturo Magidin
    Apr 26, 2012 at 0:37
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    $\begingroup$ @TundeBaba: (cont) and check to make sure that $\psi(\alpha p)$ and $\alpha\psi(p)$ are the same thing. To find a basis for the kernel, we first need to be able to recognize when something is in the kernel, which means figuring out when we have $\psi(p)=0$; to find a basis for the image, we need a generating set for the image; the values of a basis are always a generating set for the image, so figuring out $\psi(1)$, $\psi(x)$, $\psi(x^2)$, and $\psi(x^4)$ will give us a generating set for the image. Do you know how to take a generating set and pare it down to a basis? $\endgroup$
    – Arturo Magidin
    Apr 26, 2012 at 0:38

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