# I need to calculate $x^{50}$ [duplicate]

$x=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$, I need to calculate $x^{50}$

Could anyone tell me how to proceed?

Thank you.

## marked as duplicate by user1551, Git Gud, Shuhao Cao, TZakrevskiy, lhfAug 28 '13 at 14:44

• – GEL Aug 27 '13 at 14:33
• in particular, $x^{50} = x^{32}x^{16}x^2$, so you can do it with only 7 matrix multiplications. – Carl Mummert Aug 27 '13 at 14:38
• Do the first few powers by hand. There is a pattern that you should spot. – Rhys Aug 27 '13 at 14:40
• There is a pattern, just work out a few powers. – dajoker Aug 27 '13 at 14:41
• Is this not the Cayley Hamilton theorem? (Matrix of eigenvectors * matrix of eigenvalues * matrix of eigenvectors inverted = matrix, so Matrix^n is just matrix of eigenvalues ^n which only has entries along the diagonal?) – Alec Teal Aug 27 '13 at 17:34

Start by computing $X^2 = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}$.

Since $X^2 = I + N$ where $N^2 = 0$, and what you ask for is $(X^2)^{25} = I^{25} + 25I^{24}N + (\ldots)N^2$, the answer may be found by inspection:

$$X^{50} = I + 25N$$

This is a very elementary approach based on finding the general form. If we do $x^2$, we find $$x^2=\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\end{pmatrix},~~x^4=\begin{pmatrix}1&0&0\\2&1&0\\2&0&1\end{pmatrix}$$ so I guess that we have $$x^{2k}=\begin{pmatrix}1&0&0\\k&1&0\\k&0&1\end{pmatrix}$$ An inductive approach adimits this general form is valid for integers $k>0$.

The Jordan Decomposition yields $$\left[ \begin{array}{r} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right] = \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{r} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right]^{-1}$$ Block matrices are easier to raise to a power: \begin{align} \left[ \begin{array}{r} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]^{50} &= \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{r} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right]^{50} \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right]^{-1}\\[6pt] &= \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right] \left[ \begin{array}{r} 1 & 0 & 0 \\ 0 & 1 & 50 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{r} 0 & 0 & 2 \\ -1 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right]^{-1}\\[6pt] &= \left[ \begin{array}{r} 1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1 \end{array} \right] \end{align}

The graph with adjacency matrix $$x:=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$$ is drawn below:

For $i,j \in \{1,2,3\}$, the element in cell $(i,j)$ in $x^{50}$ is the number of walks from vertex $i$ to vertex $j$ of length $50$ in the above graph.

We can thus immediately deduce from the graph structure that $x^{50}$ has the form $$x^{50}:=\begin{pmatrix}1&0&0\\?&1&0\\?&0&1\end{pmatrix}$$ since $50$ is even.

With a bit of thought, we can surmise that there are $25$ walks from vertex $2$ to vertex $1$ of length $50$ (we can land at vertex $1$ after step $1,3,\ldots,49$, after which the walk is determined). Similarly, there are $25$ walks from vertex $3$ to vertex $1$ of length $50$.

Hence $$x^{50}:=\begin{pmatrix}1&0&0\\25&1&0\\25&0&1\end{pmatrix}.$$

• What is the rationale behind representing a matrix multiplication problem as a graph traversal problem? What is the intuition behind that and how did you get it? – Ali Dec 7 '16 at 16:03
• Great answer! For me a completely new idea how to approach such a problem. – Elmar Zander Nov 13 '18 at 8:59

Evaluate the first few powers and guess $\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}^{2n}$ = $\begin{pmatrix}1&0&0\\n&1&0\\n&0&1\end{pmatrix}$, proof by induction. Then $x^{50}=\begin{pmatrix}1&0&0\\25&1&0\\25&0&1\end{pmatrix}$.

Try to prove by induction that: $$A^n=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}^n = \begin{pmatrix}1&0&0\\ \left\lceil\frac{n}{2}\right\rceil&n+1\bmod{2}&n\bmod{2}\\\left\lfloor\frac{n}{2}\right\rfloor&n\bmod{2}&n+1\bmod2\end{pmatrix}$$

It helps if you start off by calculating a few of the powers by hand, e.g. the first four. We have $$M^2 = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$$ $$M^3 = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$$ $$M^4 = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 2 & 0 & 1 \end{array}\right]$$ My eye picks out a pattern in $M^2$ and $M^4$. Can you prove by induction that $$M^{2n} = \left[\begin{array}{ccc} 1 & 0 & 0 \\ n & 1 & 0 \\ n & 0 & 1 \end{array}\right]$$ Check the case for when $n=1$, then assume that it works for $n=k$, and show that this implies it also holds for $n=k+1$. You should be able to see the final answer.

Try to diagonalize the matrix. Say we could write $X=PDP^{-1}$ where $D$ is diagonal. Then computation would be easy since $X^{50}=PD^{50}P^{-1}$.

• The matrix isn't diagonalizable. – David Mitra Aug 27 '13 at 14:38
• The substitute for diagonalization is the Jordan normal form. Then the 50th power is easy to compute. – GEdgar Aug 27 '13 at 14:40
• stupid question: why isn't it diagonalizable? The det = -1, which means it should have 3 unique eigenvalues/vectors, right? – user60462 Nov 2 '15 at 4:27
• @user60462 For a matrix $A$, $\det A \neq 0$ means it is invertible but is not directly related to diagonalization (there are diagonalizable singular matrices, such as any symmetric positive semi-definite matrix, and non-diagonalizable invertible matrices such as this one). A $n\times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. In this case, the eigenvalue $1$ has a multiplicity 2 but the corresponding eigenspace is only 1 dimensional therefore $A$ is not diagonalizable. – jld Jan 16 '17 at 1:34