Linear algebra: Find the square roots of 8 and 10 in F11 (integers mod 11)

[Ryker]

Homework Statement

Find the square roots of 8 and 10 in $$\mathbb{F}_{11}$$, ie. the integers mod 11.

The Attempt at a Solution

Well, I ran over all the possibilities, but I can't find it. It has been suggested to me that perhaps you could express it in terms of other square roots somehow, but I am lost finding it. Any suggestions?

 

[micromass]

Well, it appears that 8 and 10 have no square roots in $$\mathbb{F}_{11}$$. Have you covered field extensions??

 

[Ryker]

I don't think we have, or perhaps we might have done some stuff pertaining to them, but not actually calling them field extension (we've done, for example, direct sums, quotient spaces etc.). But would then one be able to find the square root of those numbers in an extended field? Also, this is part of a diagonalisation question, so would finding a root in an extended field change the applicability of the solution?

 

[micromass]

Can you perhaps post the entire questions, because this is impossible to do without field extensions.

Basically, what I am getting at is: extend the field $$\mathbb{F}_{11}$$ with a number i which satisfies $$i^2=-1$$ (mod 11). You then obtain a field $$\mathbb{F}_{11}$$ whose numbers are of the form a+bi with $$a,b\in \mathbb{F}_{11}$$. In the extended field $$\mathbb{F}_{11}$$, one can find square roots of 8 and 10. For example: the square root of 10 is simply i.

 

[Ryker]

I'd rather not post the entire question just to be on the safe side (although I'm basically posting the question, because I think there might've been an error somewhere in the problem itself), but we were given a matrix T and asked to diagonalise it. I've done that, double checked the solution (also on Wolfram Alpha, just to be sure there isn't a mistake on my part there), and then we are to use that diagonalized matrix in order to find the square root of T, knowing that

$$D = X^{-1}TX$$,

where D is the diagonal matrix, and X is the matrix of eigenvectors. So

$$\sqrt T = X \sqrt D X^{-1}$$,

and since D is diagonal, the entries in its square root are just square roots of entries in D.

But I do see that the question asks that in our solution for square roots of T the entries should be numbers in $$\mathbb{F}_{11}$$ (ie. numbers from the list 0, 1, ..., 10), so it seems that field extension then couldn't be the answer, right?

 

[micromass]

Well, I see no other answer. The only way you can find a square root of 8 and 10, is by a field extension...

 

[Ryker]

Alright, thanks for the help, I'm glad it's not just me not being crafty enough