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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?

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

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    $\begingroup$ Yes. This idea can be conceived by noting that the given matrix is upper triangular, then maybe an upper triangular solution $A$ is possible, and the eigenvalues of the original matrix are $1$ and $1$, then maybe the eigenvalues of $A$ can be taken to be also $1$ and $1$ (a particular tenth root of $1$ is $1$). So if you try it out, you will succeed. A more difficult problem is to find all solutions $A$ with entries in some field (such as $\mathbb{Q}$ or $\mathbb{C}$). $\endgroup$ – Jeppe Stig Nielsen Jan 29 '18 at 18:08
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    $\begingroup$ @JeppeStigNielsen Indeed, that's a more difficult problem. $\endgroup$ – José Carlos Santos Jan 29 '18 at 19:10
  • $\begingroup$ (+1) This is the approach that first came to mind. Now I need to approach this slightly differently. $\endgroup$ – robjohn♦ Jan 30 '18 at 18:10
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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 $k>1$.

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    $\begingroup$ Note that it is usually not possible to use the binomial formula for matrices, because the product is not commutative. In this case it is fine, because one of the matrices is the identity. $\endgroup$ – M.Herzkamp Jan 29 '18 at 11:45
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    $\begingroup$ @M.Herzkamp Yes, we can use the formula because matrices commute. Thank you for valuable remark.. $\endgroup$ – Widawensen Jan 29 '18 at 11:49
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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 $$A^n = \left(QD^{1/n}Q^{-1}\right)^n = Q\left(D^{1/n}\right)^nQ^{-1} = QDQ^{-1} = M$$

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: $$\begin{bmatrix}1 & 1\\0&1\end{bmatrix}$$ is not diagonalizable.

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  • $\begingroup$ If you can figure out how exponentials of Jordan blocks work, you can use Jordan normal form instead of just diagonalization. $\endgroup$ – Kyle Miller Jan 30 '18 at 5:52
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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):

enter image description here

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).

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  • $\begingroup$ Is that supposed to be $f'(x)=10x^9$? $\endgroup$ – ASKASK Jan 30 '18 at 11:52
  • $\begingroup$ @ASKASK yep must have gotten an elf in my computer who ate it away ;) $\endgroup$ – mathreadler Jan 30 '18 at 11:53
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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 $A=a\pmatrix{1&c\\0&1}$ for some $c$. We immediately get $c=1/10$ (as long as $10$ is invertible in your field).

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

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