-3

let
$H=\{ $$ \begin{bmatrix} a&-b\\ b&a \end{bmatrix} \in M_2(R)$$|a,b \in R\}$

and $$M_2(R)$$ is the sets of all 2*2 matrices and all entries in them are Real number.
let
$$f:\mathbb C \to H ; f(a+bi) = \begin{bmatrix} a&-b\\ b&a \end{bmatrix} $$

how can I prove that f is onto? my solution is like this:
for every $$h\in H$$ there must be a $$c\in \mathbb C$$ where f(c)= h.
we assume that there exist an h that is not correct in this condition.
$$ h = \begin{bmatrix} a&-b\\ b&a \end{bmatrix} $$ so we can say there is a c=a+bi.
But I think this solution is not correct so what should I do?

Aidayg24
  • 5
  • In fact $f$ is bijective (one-to-one and onto): For any complex number $z \in \mathbb{C}$, there exists a unique matrix $h \in H$, and vice versa. (And $f$ is a field isomorphism: You can add and multiply matrices in $H$ that mirror the corresponding addition and multiplication in $\mathbb{C}$.) – Brian Tung Feb 25 '22 at 18:58
  • @JoséCarlosSantos Yes! thanks it was really helpful. – Aidayg24 Feb 25 '22 at 19:10
  • @BrianTung Yes you are right. thank – Aidayg24 Feb 25 '22 at 19:11

2 Answers2

3

Choose an arbitrary member

$ h= \left[ {\begin{array}{cc} a' & -b' \\ -b' & a' \\ \end{array} } \right] $

of the codomain set $H$. Then $a',b'\in{}\mathbb{R}$.

The function $f$ maps $(a'+b'i)$ to $h$. Because $h$ was arbitrary and we found a domain element which $f$ maps to $h$, $f$ is onto.

Gary
  • 515
0

I think you've got it, mostly. The fact that $f$ is onto is not complicated, and it's mostly given away by the definition of $f$:

$$f(a + ib) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}.$$

For any $h \in H$ (remember: $H$ is the codomain), we know $h$ takes a particular form: there exist $a$ and $b$ such that $$h = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}.$$ This is what it means for $h$ to belong to the set $H$!

We need to find a complex number $z$ such that $f(z) = h$. Let's just choose $z = a + ib$. Then, $$f(z) = f(a + ib) = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} = h$$ by the definition of $f$. This completes the proof that $f$ is onto.

Think of it like a recipe. How do we turn matrices $H$ into the complex numbers that map to them? Which complex number maps to $\begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}$? Which complex number maps to $\begin{bmatrix} -3 & 4 \\ -4 & -3 \end{bmatrix}$? Which complex number maps to $\begin{bmatrix} e & -\pi \\ \pi & e \end{bmatrix}$? All you're supposed to do is put to paper whatever process you use to find these complex numbers. As it turns out, it's quite an easy process!

Theo Bendit
  • 50,900