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Let $T: P_2(\mathbb{F}_4) \rightarrow \mathbb{F}_4^2$ be given by: $T(eX^2+fX+g)=(e+\alpha f, (1+\alpha)g)$, where $\mathbb{F}_4 = \{0,1,\alpha, 1 + \alpha\}$ with $\alpha^2+\alpha+1=0$. Find with justification: a) $[T]_A^B$, where $A$ and $B$ are standard bases for $P_2(\mathbb{F}_4)$ and $\mathbb{F}_4^2$, respectively, b) a complete list of $ker(T)$ and $range(T)$.

So far I have plugged in values from A into the function and have a matrix looking like this: \begin{bmatrix} 0 & 0 & \alpha & 1+\alpha \\ 0 & 1+\alpha & 0 & 1+\alpha \end{bmatrix}

Is this enough for part a? From here I should be able to get part b.

Let me know if I am on the right track or if I am doing something wrong. Thanks in advance.

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    How do you think of $\mathbb{F}_4$? In the usual notation, $(e+\alpha f, (1+\alpha)g)\not\in\mathbb{F}_4$...2012-09-26

1 Answers 1

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I guess, this $T$ transformation goes from the set of polynomials over $F_4$ of degree at most 2 to ${F_4}^2$, the 2d space over $F_4$, no?

What values have you plugged in? You need to apply $T$ on the elements of the given basis $A$. It says, the standard basis, on $P_2(F_4)$, that is the 3 polynomials: $1,x,x^2$. Ok, so calculate $T(1)$, $T(x)$ and $T(x^2)$, these will be the columns of the matrix. So, this is going to be a $3\times 2$ matrix.

For b) you don't really need the matrix. Try to show that any 2d vector is in the range of $T$ (for this, you don't need the fact that the field is $F_4$), and calculate which $(e,f,g)$ coefficients give $0$ when applied $T$.

Let $\beta:=\alpha+1$.

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    Sorry about that I did mean $\mathbb{F}_4^2$. I plugged in the values of $\mathbb{F}_4$ into the polynomial expression, but what you have said makes more sense.2012-09-27
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    So for this a) I get that the matrix is \begin{bmatrix} 0 & 1+\alpha \\ \alpha & 0 \\ 1 & 0 \end{bmatrix} Is that correct?2012-09-27
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    Well, it's usually meant the transpose: the results written in the *columns* (because the matrix multiplication $A\cdot B$ corresponds to the composition of linear maps $A\circ B$): $$\begin{bmatrix} 0 & \alpha & 1 \\ 1+\alpha & 0 & 0 \end{bmatrix}$$ However, this all could be turned around, if composition would be meant from left to right (but that's not usual in this context)..2012-09-27
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    So to get the range(T) I would do the following: \begin{bmatrix} 0 & \alpha & 1 & e+\alpha f \\ 1+\alpha & 0 & 0 &(1+\alpha)g \end{bmatrix}. And I would see that this would require that $e=f=g=1. Am I going about this correctly?2012-09-27
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    I still do not know how to get the range(T) or ker(T) from this.2012-09-27