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Determining if a function is onto and one to one?

f --R^(2x2) --> R
f(a b) (that's meant to be a matric) = ad - bc
(c d)

Frist a found an easy counter example for one to one i.e. a = -a etc. However I'm totally stuck in determining if it's onto. Usually I'd try and find an inverse but I've got no clue how to do that.
Original post by Roger The Doger
f --R^(2x2) --> R
f(a b) (that's meant to be a matric) = ad - bc
(c d)

Frist a found an easy counter example for one to one i.e. a = -a etc. However I'm totally stuck in determining if it's onto. Usually I'd try and find an inverse but I've got no clue how to do that.


f:R2×2→R f:\mathbb{R}^{2 \times 2} \rightarrow \mathbb{R}
f((abcd))=ad−bcf\begin{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix}\end{pmatrix}=ad-bc
Yes your right for 1-1 counterexample could be f((1001))f\begin{pmatrix}\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}\end{pmatrix}=f((−100−1))=1=f\begin{pmatrix}\begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}\end{pmatrix}=1

Remember onto means the image of f is equal to the codomain: in this case it means you can construct a matrix (abcd)\begin{pmatrix} a & b \\c & d \end{pmatrix} such that f((abcd))=rf\begin{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix}\end{pmatrix}=r ∀r∈R \forall r \in \mathbb{R}. An example (a001)\begin{pmatrix} a & 0 \\0 & 1 \end{pmatrix} notice that f((a001))=af\begin{pmatrix}\begin{pmatrix} a & 0 \\0 & 1 \end{pmatrix}\end{pmatrix}=a ∀a∈R \forall a\in \mathbb{R} so the image equals the codomain.
Original post by tombayes
f:R2×2→R f:\mathbb{R}^{2 \times 2} \rightarrow \mathbb{R}
f((abcd))=ad−bcf\begin{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix}\end{pmatrix}=ad-bc
Yes your right for 1-1 counterexample could be f((1001))f\begin{pmatrix}\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}\end{pmatrix}=f((−100−1))=1=f\begin{pmatrix}\begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}\end{pmatrix}=1

Remember onto means the image of f is equal to the codomain: in this case it means you can construct a matrix (abcd)\begin{pmatrix} a & b \\c & d \end{pmatrix} such that f((abcd))=rf\begin{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix}\end{pmatrix}=r ∀r∈R \forall r \in \mathbb{R}. An example (a001)\begin{pmatrix} a & 0 \\0 & 1 \end{pmatrix} notice that f((a001))=af\begin{pmatrix}\begin{pmatrix} a & 0 \\0 & 1 \end{pmatrix}\end{pmatrix}=a ∀a∈R \forall a\in \mathbb{R} so the image equals the codomain.


Thanks you! Had been stuck on that for a while happen thought about putting it like you did, makes so much more sense that the stuff I tried.

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