Ytukyg

I want to find a $2 times 2$ matrix, named $A$ in this situation, such that:

$$A^{10}=begin {bmatrix} 1 & 1 \ 0 & 1 end {bmatrix} $$

How can I get started? I was thinking about filling $A$ with arbitrary values $a, b, c, d$ and then multiplying it by itself ten times, then setting those values equal to the given values but I quickly realized that would take too long. Is there a more efficient way?

Take$$A=begin{bmatrix}1&x\0&1end{bmatrix}.$$Now, compute $A^2,A^3,ldots$ You'll find quickly which $x$ you should choose.

You can also consider $$A=begin {bmatrix} 1 & 0 \ 0 & 1 end {bmatrix} + begin {bmatrix} 0 & a \ 0 & 0 end {bmatrix} =I+N $$ and try to use binomial formula for expansion of powers of a binomial.

Notice that $N^k=0$ for $k>1$.

The answers given by José Carlos Santos and Widawensen are correct. There is a more general alternative method, even through it does not work for your particular example. But I thought I'd give it anyway in case it is useful to others.

If you want the $n^{text{th}}$ root of the matrix $M$, and $M$ is diagonalizable - i.e., there exists $Q$ such that $D = Q^{-1}MQ$ is a diagonal matrix, then $D^{1/n}$ is just the matrix whose diagonal elements are those of $D$ raised to the $1/n$ power. So you can set $$A = QD^{1/n}Q^{-1}$$ and find that $$A^n = left(QD^{1/n}Q^{-1}right)^n = Qleft(D^{1/n}right)^nQ^{-1} = QDQ^{-1} = M$$

To diagonalize $M$, you find it's eigenvalues and their corresponding eigenvectors. If the eigenvectors span the entire space, then $M$ is diagonalizable, the columns of $Q$ are eigenvectors of $M$, and the diagonal elements of $D$ are the eigenvalues.

Alas, it is not applicable here: $$begin{bmatrix}1 & 1\0&1end{bmatrix}$$
is not diagonalizable.

Ok this may seem a bit overkill for this particular question, but please bear with me as it could help others.

You can use the Newton-Rhapson method, trying to solve the equation

$$f(x) = x^{10}-d=0$$
where
$$f'(x) = 10x^9$$
and $d$ is that matrix of yours and the iteration:

$${bf X_{n+1}} ={bf X_n}- f({bf X_n})f'({bf X_n})^{-1}$$
As long as you set initial $bf X_0$ to nothing too crazy, it will work. In fact, this is the simplest example of an discrete fractional integration solution, if you look at larger matrices $d$ filled with 1 on and below the diagonal and 0 otherwise (ok, transpose of the same matrix but same idea).

The fractional integral operators in fractional calculus found if guessing at the $bf X_0 = bf I$ matrix (described as linear convolutional filters, corresponding to a row in the solution matrix):

The two we recognize is the one in the middle (constant, corresponding to normal integral) and the one at the end (linear, corresponding to double integration).

Such a matrix $A$ will commute with $pmatrix{1&1\0&1}$, and so be of the form $pmatrix{a&b\0&a}$. This means that $a^{10}=1$ and so
$A=apmatrix{1&c\0&1}$ for some $c$. We immediately get $c=1/10$
(as long as $10$ is invertible in your field).

Hint: Another approach is to note that
$$
expleft(begin{bmatrix}0&x\0&0end{bmatrix}right)=begin{bmatrix}1&x\0&1end{bmatrix}
$$