Question

The map f : R[x] -> M(subscript3)(R) is defined by
f ( a(subscript 0) + a(subscript 1)x + a(subscript 2)x(power 2) + ....... + a(subscript n)x(power n)
_ _
| a(subscript 0) a(subscript 1) a(subscript 2) |
= | 0 a(subscript 0) a (subscript 1) |
|_ 0 0 a(subscript 0) _|
Show that f is a group homomorphism.Determine ker(f) also.

Answer 1

Answer on Question #45471 – Math - Abstract Algebra

Problem.

The map $f: \mathbb{R}[\mathbf{x}] \to \mathbb{M}(\text{subscript}3)(R)$ is defined by

$$f \left(a (\text{subscript} 0) + a (\text{subscript} 1) x + a (\text{subscript} 2) x (\text{power} 2) + \dots + a (\text{subscript} n) x (\text{power} n) \right)$$ $$\begin{array}{l} \neg a (\text{subscript} 0) a (\text{subscript} 1) a (\text{subscript} 2) | \\ = | 0 a (\text{subscript} 0) a (\text{subscript} 1) | \\ | \_ 0 0 a (\text{subscript} 0) \_ | \\ \end{array}$$

Show that $f$ is a group homomorphism. Determine $\ker(f)$ also.

Solution.

Let $P(x) = a_0 + a_1x + \dots + a_nx^n$ and $Q(x) = b_0 + b_1x + \dots + b_nx^n$. Then

$$f \big (P (x) \big) f \big (Q (x) \big) = \left[ \begin{array}{c c c} a _ {0} & a _ {1} & a _ {2} \\ 0 & a _ {0} & a _ {1} \\ 0 & 0 & a _ {0} \end{array} \right] \left[ \begin{array}{c c c} b _ {0} & b _ {1} & b _ {2} \\ 0 & b _ {0} & b _ {1} \\ 0 & 0 & b _ {0} \end{array} \right] = \left[ \begin{array}{c c c} a _ {0} b _ {0} & a _ {0} b _ {1} + a _ {1} b _ {0} & a _ {0} b _ {2} + a _ {1} b _ {1} + a _ {2} b _ {0} \\ 0 & a _ {0} b _ {0} & a _ {0} b _ {1} + a _ {1} b _ {0} \\ 0 & 0 & a _ {0} b _ {0} \end{array} \right]$$

and

$$\begin{array}{l} f \big (P (x) Q (x) \big) = f \left(a _ {0} b _ {0} + \left(a _ {0} b _ {1} + a _ {1} b _ {0}\right) x + \left(a _ {0} b _ {2} + a _ {1} b _ {1} + a _ {2} b _ {0}\right) x ^ {2} + \dots\right) \\ = \left[ \begin{array}{c c c} a _ {0} b _ {0} & a _ {0} b _ {1} + a _ {1} b _ {0} & a _ {0} b _ {2} + a _ {1} b _ {1} + a _ {2} b _ {0} \\ 0 & a _ {0} b _ {0} & a _ {0} b _ {1} + a _ {1} b _ {0} \\ 0 & 0 & a _ {0} b _ {0} \end{array} \right]. \end{array}$$

Hence $f\big(P(x)\big)f\big(Q(x)\big) = f\big(P(x)Q(x)\big)$.

Then $f$ is group homomorphism.

$$\begin{array}{l} \ker f = \left\{P \in \mathbb {R} [ x ] \middle | f (P) = \left[ \begin{array}{c c c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \right\} = \left\{a _ {0} + a _ {1} x + \dots + a _ {n} x ^ {n} \in \mathbb {R} [ x ] | a _ {0} = a _ {1} = a _ {2} = 0 \right\}. \\ \ker f = \left\{a _ {3} x ^ {3} + \dots + a _ {n} x ^ {n} \mid a _ {i} \in \mathbb {R}, i = 3.. n \right\}. \end{array}$$

Answer: $\ker f = \{a_3x^3 + \dots + a_nx^n \mid a_i \in \mathbb{R}, i = 3..n\}$.

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