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

##### 4 Answers

#### Explanation:

By Cayley-Hamilton, the

I get

#### Explanation:

I used

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

The only real solution is

From equation 1 (0r eq 4) we now get

At this point, I can find no other restrictions on

So, I tried

Unless I made some silly error,

More generally,

If

then

Now if we add

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

First consider matrices of the form

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

In other words, such matrices form a *field*. This field is *isomorphic* to

#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

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

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))#

**Footnote**

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.

One solution is the companion matrix

#### Explanation:

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

**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

#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^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))#

**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))#