Find the nth power of a matrix

Question

Let matrix A is

$$A=\left (\begin{array}{rrr} 1&0&0 \\ 1&1&0\\ 0&0&1 \end{array}\right)$$

How can I find the 30 th power of A.. Is diagonalization possible?

I found the eigen values But cannot continue.

Answer 1

You can try to calculate the first terms and see if there's a pattern. $$ A^2 = \begin{pmatrix} 1&0&0\\ 2&1&0\\ 0&0&1 \end{pmatrix} $$ and $$ A^3= \begin{pmatrix} 1&0&0\\ 3&1&0\\ 0&0&1 \end{pmatrix}. $$

Maybe you should try to prove using induction that $$ A^{n}=\begin{pmatrix} 1&0&0\\ n&1&0\\ 0&0&1 \end{pmatrix}. $$

Answer 2

Notice that $$ \begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ b & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ a+b & 1 \end{pmatrix} \quad \text{for all $a,b \in K$}, $$ where $K$ denotes the ground field. Therefore $$ \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix}. $$ Now notice that your matrix is a block matrix of the form $$ \begin{pmatrix} 1 & 0 & \\ 1 & 1 & \\ & & 1 \end{pmatrix} = \begin{pmatrix} A & \\ & B \end{pmatrix} $$ with $$ A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \quad\text{and}\quad B = (1). $$

Therefore $$ \begin{pmatrix} 1 & 0 & \\ 1 & 1 & \\ & & 1 \end{pmatrix}^n = \begin{pmatrix} A & \\ & B \end{pmatrix}^n = \begin{pmatrix} A^n & \\ & B^n \end{pmatrix} = \begin{pmatrix} 1 & 0 & \\ n & 1 & \\ & & 1 \end{pmatrix}. $$

PS: Alternatively one can realize that your matrix is an elementary matrix, namely the one adding the fist row to the second. Applying this $n$-times is the same as just adding the first row $n$-times to the second, which is represented by the matrix we calculated.