# Proving a matrix is Bijective

1. Nov 24, 2009

### Ubernisation

1. The problem statement, all variables and given/known data

C=
a -b
b a
and p is the mapping Complex numbers --> C by the rule p(a+ib)=
a -b
b a
Show that p is a bijection

2. Relevant equations

3. The attempt at a solution

Well I need to prove it is both injective and surjective. For Injectivity I have tried showing that the determinate=0, but all I got was detA=a^2 + b^2 and for surjectivity. I also know that any non square matrix cannot be bijective, so attempted to involve this in the answer, but I don't think this is the right thing to be doing.

I have no idea what to do. I have not tackled questions similar to this before, so don't really have any idea what path I should be following.

Any help hugely welcome.

Last edited: Nov 24, 2009
2. Nov 24, 2009

### Staff: Mentor

Your problem description is confusing. Are you saying that p maps a complex number to a 2 x 2 matrix? In any case, this is a square matrix, so don't waste your time talking about nonsquare matrices.

To show that the mapping is injective (which if I recall is a synonym for one-to-one), the determinant can't be zero. You want to show that distinct complex numbers map to distinct matrices. I.e., if a1 + b1i $\neq$ a2 + b2i, then p(a1 + b1i) $\neq$ p(a2 + b2i). Equivalently, if p(a1 + b1i) = p(a2 + b2i), then a1 + b1i = a2 + b2i.

To show that the mapping is surjective (onto), show that for any 2 x 2 matrix of the form in this problem, there is a complex number a + bi such that p(a + bi) = that matrix.

3. Nov 24, 2009

### HallsofIvy

Staff Emeritus
Also, as to your title "proving a matrix is bijective", you don't want to prove anything about "bijective matrices". I think that mislead Marl44. A matrix represents a linear transformation and the linear transformation represented by a square matrix is bijective if and only if the determinant of the matrix is non-zero. There is no such condition on the determinants of the matrices here. For example, what matrix is the complex number 0 mapped to by this mapping? What is the determinant of that matrix?

What you want to prove is that the mapping
$$a+ bi \rightarrow \begin{bmatrix}a & -b \\ b & a\end{bmatrix}$$
is a bijective mapping from C, the set of complex numbers, to the set of two by two anti-symmetric matrices is bijective.

To prove it is "injective" (one-to-one) You need to prove that if a+ bi and c+ di are mapped to the same matrix, then a+ bi= c+ di. That should be easy. To prove it is surjective, you need to show that some a+ bi is mapped into any matrix of the form
[tex]\begin{bmatrix}x & -y \\ y & x\end{bmatrix}[/itex]
and you can do that by showing what a and b must equal to give that.

Last edited: Nov 24, 2009