Question

How do you find a `2xx2` matrix `A` with rational coefficients such that `A^2+A+((1,0),(0,1)) = ((0,0),(0,0))` ?

Answer 1

`A=((-1/2 - (i sqrt[3])/2,0),(0,-1/2+ (i sqrt[3])/2))`

Explanation:

By Cayley-Hamilton, the `A` characteristic polynomial is given by

`p(lambda) = lambda^2+lambda+1=0` and the eigenvalues are

`lambda=-1/2 pm (i sqrt[3])/2` so the matrix

`A=((-1/2 - (i sqrt[3])/2,0),(0,-1/2+ (i sqrt[3])/2))` is a solution.

Answer 2

I get `A = ((-1/2,b),(c,-1/2))` with `b,c in QQ` and `bc=-3/4`

Explanation:

I used `A = ((a,b),(c,d))`

I get the system (equations numbered in the obvious way)

`a^2+bc+a=-1`
`ab+bd+b=0`
`ac+cd+c=0`
`d^2+bc+d=-1`

Subtracting equation 4 from equation 1 gets us

`a^2+a-(d^2+d) = 0`

`a^2+a+1/4 -(d^2+d+1/4) = 0`

`(a+1/2)^2-(d+1/2)^2 = 0`

The only real solution is `a=d=-1/2`.

From equation 1 (0r eq 4) we now get `bc=-3/4`

At this point, I can find no other restrictions on `b` and `c`.

So, I tried `b=1` and `c = -3/4`

Unless I made some silly error,

`A = ((-1/2,1),(-3/4,-1/2))` is a solution.

More generally,

If

`A = ((-1/2,b),(c,-1/2))` with `bc=-3/4`,

then

`A^2 = ((-1/2,b),(c,-1/2))((-1/2,b),(c,-1/2)) = ((-1/2,-b),(-c,-1/2))`.

Now if we add `A`, we get

`A^2+A = ((-1,0),(0,-1))`.

Answer 3

A couple of possible solutions:

`A = ((-1/2, 1/2),(-3/2, -1/2))" "` or `" "A = ((-1/2, -1/2),(3/2, -1/2))`

Explanation:

Just in case you have not encountered the notation, the set of all rational numbers is denoted by `QQ`.

First consider matrices of the form `((a, 0), (0, a))` where `a in QQ`

If you add, subtract, multiply or divide (i.e. multiply by the multiplicative inverse), such matrices then you end up with a matrix of the same form.

Matrices of the form `((a, 0),(0, a))` where `a in QQ` are said to be closed under addition, multiplication, subtraction and multiplicative inverse (of non-zero elements). There is also an additive identity `((0, 0),(0,0))` and multiplicative identity `((1,0),(0,1))`. Both addition and multiplication are commutative.

In other words, such matrices form a field. This field is isomorphic to `QQ`, that is there is an exact correspondence between `QQ` and this field of matrices, given by the mapping:

`a -> ((a,0),(0,a))`

which preserves the arithmetical structure.

Now consider the matrix:

`((0, 1),(-3, 0))`

See what happens if we square this matrix:

`((0, 1),(-3, 0))((0, 1),(-3, 0)) = ((-3,0),(0,-3))`

So `((0, 1),(-3, 0))` is a "square root of `-3`"

If we add this to our field of matrices, and add all other matrices required to make it closed under arithmetic operations, then we have the set of matrices of the form:

`((a, b),(-3b, a))" "` where `a, b in QQ`

This corresponds to the Complex number `a + bsqrt(-3)`, behaving arithmetically just like it.

Now:

`x^2+x+1 = (x+1/2)^2-(sqrt(-3)/2)^2`

`color(white)(x^2+x+1) = (x+1/2-sqrt(-3)/2)(x+1/2+sqrt(-3)/2)`

Hence zeros:

`x = -1/2+-sqrt(-3)/2`

which correspond to the matrices:

`((-1/2, 1/2),(-3/2, -1/2))" "` and `" "((-1/2, -1/2),(3/2, -1/2))`

`color(white)()`
Footnote

`((0, 1),(-3, 0))` is only one of a family of matrices with rational coefficients with square `((-3, 0),(0, -3))`.

In fact any matrix of the following form will work:

`((0, k),(-3/k, 0))`

and hence there are a family of solutions to the original problem.

Answer 4

One solution is the companion matrix `((0, -1),(1, -1))`

Explanation:

I think I was taught this about 35 years ago and had forgotten it...

`color(white)()`
Companion matrix

Given a monic polynomial:

`x^n + a_(n-1)x^(n-1) + a_(n-2)x^(n-2) + ... + a_1 x + a_0`

The companion matrix is an `n xx n` matrix of the form:

`C = ((0, 0,..., 0, -a_0), (1, 0,..., 0, -a_1), (0, 1,...,0, -a_2), (vdots, vdots,ddots,vdots,vdots),(0, 0,...,1, -a_(n-1)))`

Then `C` satisfies:

`C^n + a_(n-1)C^(n-1) + a_(n-2)C^(n-2) + ... + a_1 C + a_0 = 0`

For a monic quadratic polynomial:

`x^2+bx+c`

The companion matrix is:

`C = ((0, -c), (1, -b))`

So for our example:

`x^2+x+1`

`C = ((0, -1), (1, -1))`

`color(white)()`
Footnote

I was interested in more complicated examples such as:

`x^5+4x+2 = 0`

The companion matrix for this example is:

`((0, 0, 0, 0, -2), (1, 0, 0, 0, -4), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0))`