Finding g(t) for Characteristic Polynomial f(t) = t2 - 5t + 4 and Matrix A

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Homework Statement


Suppose A is a 2x2 real matrix with characteristic polynomial f(t) = t2 - 5t +4. Find a real polynomial g(t) of degree 1 such that (g(A))2 = A.

Suppose A is a 2x2 complex matrix with A2 ≠ O. Show that there is a complex polynomial g(t) of degree 1 such that (g(A))2 = A.

Homework Equations


Cayley-Hamilton Theorem


The Attempt at a Solution


I found that det A = 4 and from f(A) = 0, I found the inverse of A to be 1/4 * A(A-5).

I am completely stuck after this. I let A be a matrix of 4 variables a, b, c, and d, then tried to solve for the variables but ended up with really long and messy terms.

Can someone give me a lead as to what the g(t) has to do with f(t)?
 
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You know that (g(A))^2 = A, and g is linear, so why don't you write g(A) = aA + b and see what happens.
 
I tried that too before but got nowhere so I thought it was irrelevant. I get simply a quadratic for A, where 0 = a2A2 + (2ab - 1)A + c2. Does this have to do with f(t) = t2 -5t + 4?

EDIT 1:
I equated the coefficients and found that a = +/- 1 and b = -/+ 2. So then I just substitute this back into the coefficients for g(t)?

EDIT 2:
All right, I got it now, thanks. =)

Is the second question similar to the first?

EDIT 3:
I tried to find the characteristic polynomial for A with a, b, c, d variable entries, then I isolated for A, used the (forgot name) method to get a quadratic equation. Is this correct?

EDIT 4:
Never mind, I think I solved it. I isolated for A (dividing both sides by a+ d is possible because from the determinant, we get that a + d =/= 0).

A = (1/(a+d)) A2 + (ad-bc)/(a+d)
A + eA = (1/(a+d)) A2 + eA + (ad-bc)/(a+d) , e =/= -1

Then, I isolated for A again, equated it to the square of a linear equation g(t) = mt + n, found m and n in terms for the 5 variables above, isolated for e in terms of the other 4 variables, and substituted it all into g(t). The only restriction is that ad - bc =/=, but the questions asks for any complex function.

Is this all right? Thanks.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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