Show that ##\mathbb{C}## can be obtained as 2 × 2 matrices

[Karl Karlsson]

I have this problem in my book:

Show that $\mathbb{C}$ can be obtained as 2 × 2 matrices with coefficients in $\mathbb{R}$ using an arbitrary 2 × 2 matrix $J$ with a characteristic polynomial that does not contain real zeros.

In the picture below is the given solution for this:

I understand that the eigenvalues must be conjugates. But I don't understand how they so quickly arrived at the finite field extension K

The only theorem I know related to solving the problem above is:

Theorem: If A is a matrix in $M_n(k)$ and the minimal polynomial of A is irreducible, then $K = \{p(A): p (x) \in k [x]\}$ is a finite field

Thanks in advance!

 

[DrClaude]

You'll have to provide a translation of the text.

I can read it, but must others here can't

 

[PeroK]

I have this problem in my book:

Show that $\mathbb{C}$ can be obtained as 2 × 2 matrices with coefficients in $\mathbb{R}$ using an arbitrary 2 × 2 matrix $J$ with a characteristic polynomial that does not contain real zeros.

In the picture below is the given solution for this:
View attachment 270276
I understand that the eigenvalues must be conjugates. But I don't understand how they so quickly arrived at the finite field extension K

The only theorem I know related to solving the problem above is:

Theorem: If A is a matrix in $M_n(k)$ and the minimal polynomial of A is irreducible, then $K = \{p(A): p (x) \in k [x]\}$ is a finite field

Thanks in advance!

Is $J$ supposed to be the matrix:
$$J=\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}$$

 

[mathwonk]

given a complex number z, and a pair of real numbers (a,b), defining T(a,b) as z.(a+bi), and taking real and imaginary parts, gives a real linear map R^2-->R^2, which must then be representable as a 2x2 real matrix. Moreover, (z.w)(a+bi) = z.(w.(a+bi)), so there is an embedding of the ring of complex numbers into the ring of real linear transformations of R^2, i.e. 2x2 real matrices. To find the matrix representing multiplication by z = c+di, calculate its action on a real basis for R^2, namely on (1,0) and on (0,1).

edit: Pardon me, I only read the first line of your problem question. So this solves it for the matrix J with columns [0 1] and [-1 0], and char poly t^2+1.

But the complex numbers can be obtained from the reals by adjoining any non real complex number. I.e. the splitting field over the reals, of any quadratic polynomial with no real roots, is the complexes. Put another way, the field generated over R, by 1 and a+bi, with b≠0, is C.

Since by the Cayley Hamilton theorem, every matrix is a root of its characteristic polynomial, when we take the real linear combinations of the matrices I, J, where I is the identity and J is any 2x2 matrix with irreducible (over R) char poly, we are adjoining to R.I, namely to R, the roots of J. Hence we get C = R.I + R.J.

 

[Karl Karlsson]

You'll have to provide a translation of the text.

I can read it, but must others here can't

The translation of the solution goes like:

If $J$ lacks real eigenvalues, it means that the eigenvalues are a complex conjugate pair. Thus the body we get as $K = \{xI + yJ: x, y ∈ R\}$ contains the two numbers $a \pm ib$ where b is not 0. Since it is a vector space over $\mathbb{R}$ we get that $( a + ib) - (a - ib) = 2ib$ belongs to the body and thus also $1/(2b) · 2ib = i$. Therefore K is isomorphic with the complex numbers $\mathbb{C}$.

 

[DrClaude]

If $J$ lacks real eigenvalues, it means that the eigenvalues are a complex conjugate pair. Thus the body we get as $K = \{xI + yJ: x, y ∈ R\}$ contains the two numbers $a \pm ib$ where b is not 0. Since it is a vector space over $\mathbb{R}$ we get that $( a + ib) - (a - ib) = 2ib$ belongs to the body and thus also $1/(2b) · 2ib = i$. Therefore K is isomorphic with the complex numbers $\mathbb{C}$.

I think that "field" is the proper translation here instead of "body."

 

[martinbn]

The translation of the solution goes like:

If $J$ lacks real eigenvalues, it means that the eigenvalues are a complex conjugate pair. Thus the body we get as $K = \{xI + yJ: x, y ∈ R\}$ contains the two numbers $a \pm ib$ where b is not 0. Since it is a vector space over $\mathbb{R}$ we get that $( a + ib) - (a - ib) = 2ib$ belongs to the body and thus also $1/(2b) · 2ib = i$. Therefore K is isomorphic with the complex numbers $\mathbb{C}$.

What is not clear about this solution? Except that in English it is usually called a field, not a body.

 

[PeroK]

The translation of the solution goes like:

If $J$ lacks real eigenvalues, it means that the eigenvalues are a complex conjugate pair. Thus the body we get as $K = \{xI + yJ: x, y ∈ R\}$ contains the two numbers $a \pm ib$ where b is not 0. Since it is a vector space over $\mathbb{R}$ we get that $( a + ib) - (a - ib) = 2ib$ belongs to the body and thus also $1/(2b) · 2ib = i$. Therefore K is isomorphic with the complex numbers $\mathbb{C}$.

So, $J$ could be any matrix with these properties; not only the one I suggested above. Which you might want to check has these properties.

In any case, the task is to show that $K$ is isomorphic to $\mathbb C$, in terms of the field operations of addition and multiplication.

Does this make sense to you?

 

[Karl Karlsson]

So, $J$ could be any matrix with these properties; not only the one I suggested above. Which you might want to check has these properties.

In any case, the task is to show that $K$ is isomorphic to $\mathbb C$, in terms of the field operations of addition and multiplication.

Does this make sense to you?

Hi!

Yes, I see why it holds for the matrix $J=\begin{pmatrix} 0 & 1 \\-1 & 0\end{pmatrix}$ in your example but why does it hold for arbitrary 2 × 2 matrices with complex eigenvalues?

 

[PeroK]

Hi!

Yes, I see why it holds for the matrix $J=\begin{pmatrix} 0 & 1 \\-1 & 0\end{pmatrix}$ in your example but why does it hold for arbitrary 2 × 2 matrices with complex eigenvalues?

That's what you have to prove!

Do you understand how to tackle a problem like this?

 

[Karl Karlsson]

That's what you have to prove!

Do you understand how to tackle a problem like this?

I did try to calculate the determinant of some arbitrary matrix $J=\begin{pmatrix} a_{11}-\lambda & a_{12} \\a_{21} & a_{22}-\lambda\end{pmatrix}$

This did not get me anywhere...

 

[PeroK]

I did try to calculate the determinant of some arbitrary matrix $J=\begin{pmatrix} a_{11}-\lambda & a_{12} \\a_{21} & a_{22}-\lambda\end{pmatrix}$

This did not get me anywhere...

You don't need to do any of that. First, there is an obvious mapping$$\phi: \mathbb C \rightarrow K$$ defined by$$\phi(x + iy) = xI + yJ$$
What you need to show is that this is a field isomorphism, which has properties:

1) $\phi$ must be onto (surjective).
2) $\phi$ must be one-to-one (injective).
And, for all $z_1, z_2 \in \mathbb C$:
3) $\phi (z_1 + z_2) = \phi(z_1) + \phi(z_2)$
4) $\phi(z_1z_2) = \phi(z_1)\phi(z_2)$

Have you done anything like that before?

 

[PeroK]

You don't need to do any of that. First, there is an obvious mapping$$\phi: \mathbb C \rightarrow K$$ defined by$$\phi(x + iy) = xI + yJ$$

Actually, if $J$ is arbitrary, then $\phi$ will be more complicated than that.

There's something about this that doesn't seem right to me. The field isomorphism doesn't follow (as far as I can see) from the argument in the text. Perhaps something has been lost in translation.

PS you really want $J^2 = -I$ to keep things simple.

 

[mathwonk]

(Building on my edit of post #4), you might define a map from the polynomial ring R[t] to your ring of matrices R.I+R.J, and show it is surjective with kernel the (irred. quadratic) char poly P(t) of J, and then show that R[t]/P(t) ≈ C.
I.e. R[t]/P(t) is a quadratic extension of R, and a subfield of C, QED. Notice this dimension theoretic argument finesses the question of actually writing down an isomorphism from C to R.I+R.J.
I.e. if z is any non real complex number, there is an isomorphism from R+i.R to R+z.R, but it takes a little effort to write it down explicitly.

 

[mathwonk]

The usual representation of complex numbers is as a field containing, and algebraic over, the reals, and containing a root of the equation X^2+1=0. It follows that the complex numbers are isomorphic to the quotient ring R[X]/(X^2+1). Moreover, given any commutative ring S containing a copy of the real numbers, and given any element T of S, there is a unique R algebra map R[X]-->S, taking X to T. Such a map sets up an isomorphism from R[X]/f(X) to S, where f(X) is the minimal polynomial of T.

Consequently, since the real scalar multiples of the identity matrix I, is a copy of the real numbers, and since these matrices commute with all 2x2 matrices, if we choose a matrix T such that T^2 = -I, we get a commutative subring R.I+R.T of the ring of 2x2 real matrices, and a map R[X]--> R.I+R.T, with kernel X^2+1, hence an isomorphism R[X]/(X^2+1) ≈ R.I+R.T, hence an isomorphism of the subring of matrices R.I+R.T with the complex numbers.

Now what if we choose a matrix T with minimal polynomial f(X) which is any irreducible quadratic real polynomial, i.e. any real polynomial with no real roots? Then the map R[X]-->R.I+R.T has kernel (f(T)), and defines an isomorphism R[X]/(f(T)) ≈ R.I+R.T. Thus the question boils down to showing also that R[X]/(f(X)) ≈ the complex numbers. I.e. the whole problem is the fact that R[X]/(f(X)) is isomorphic to the complex numbers when f is any irreducible quadratic real polynomial, not just X^2+1.

The argument given in the book is rather brief but boils down to assuming that the ring R[X]/(f(X)) is isomorphic to some subring of the complex numbers that contains the reals, which also contains two complex numbers of form a+bi and a-bi. They then observe that it must also contain their difference, namely 2bi, hence also it contains i, so must be the full complex numbers.

I did not see or understand the full explanation in the book, but it seemed to me somewhat short of a full explanation of a complete proof, which I have therefore tried to expand on. hope this helps. at least now i can sleep.