How to Show That the Mapping from Complex Numbers to a Matrix is Bijective

  • Thread starter Ubernisation
  • Start date
  • Tags
    Matrix
In summary, The mapping p from complex numbers to a 2 x 2 matrix is bijective if the determinant of the matrix is non-zero. To prove that p is bijective, we need to show that it is both injective and surjective. To show injectivity, we must prove that distinct complex numbers map to distinct matrices, and to show surjectivity, we must show that for any 2 x 2 matrix, there exists a complex number that maps to it. The mapping is also not affected by non-square matrices, so we do not need to consider them in our proof.
  • #1
Ubernisation
1
0

Homework Statement



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

Homework Equations


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:
Physics news on Phys.org
  • #2
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 [itex]\neq[/itex] a2 + b2i, then p(a1 + b1i) [itex]\neq[/itex] 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
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
[tex]a+ bi \rightarrow \begin{bmatrix}a & -b \\ b & a\end{bmatrix}[/tex]
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 by a moderator:

FAQ: How to Show That the Mapping from Complex Numbers to a Matrix is Bijective

What does it mean for a matrix to be bijective?

A matrix is considered bijective if it is both injective (one-to-one) and surjective (onto). This means that every element in the domain of the matrix has a unique mapping to an element in the range, and every element in the range is mapped to by at least one element in the domain.

How can I prove that a matrix is bijective?

To prove that a matrix is bijective, you can use the definition of injectivity and surjectivity. Show that for every possible input, there is a unique output, and that every possible output has at least one corresponding input. You can also use matrix operations such as row reduction to show that the matrix has a unique solution for every possible output.

Can a non-square matrix be bijective?

No, a non-square matrix cannot be bijective. Since a bijective matrix must have a unique solution for every output, a non-square matrix would have either more or less solutions than the number of outputs, thus violating either injectivity or surjectivity.

Is it possible for a matrix to be injective but not surjective (or vice versa)?

Yes, it is possible for a matrix to be injective but not surjective, or surjective but not injective. For example, a 2x2 identity matrix is both injective and surjective, while a 2x1 matrix with all zeros is injective but not surjective.

Why is proving a matrix is bijective important in mathematics?

Proving a matrix is bijective is important in mathematics because it ensures that the matrix has a unique solution for every possible output. This allows for easier manipulation and analysis of the matrix, making it a powerful tool in various fields such as linear algebra, computer science, and engineering.

Back
Top