How exactly to use Cayley's Hamilton's theorem to find $A^{50}$ in this case? (matrix recursion equation)

Question

Say $$A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ I'm supposed to find $A^{50}$ using the Cayley Hamilton theorem.

My Attempt:

$|A-\lambda I| = 0 \implies \lambda^2-2\lambda+1=0$

So $A^2-2A+I = O \implies A^{50}=2A^{49} - A^{48}$

But using this method, I'll need to know $A^{49}$ and $A^{48}$ to find $A^{50}$. But that doesn't seem the right way. This seems like some sort of matrix recursion but I'm not sure how to proceed from here.

Answer 1

Note that by the division algorithm we have that $$ x^{50} = (x^2 - 2x + 1) P(x) + ax + b $$ for some constants $a$ and $b$, and some polynomial $P(x)$.

Letting $x = 1$ gives us that $a + b = 1$. We thus have that $$ x^{49} + x^{48} + \dots + x + 1 = \frac{x^{50} - 1}{x - 1} = (x - 1) P(x) + a, $$ and substituting $x = 1$ into this relation gives us that $a = 50$, which in turn implies that $b = -49$.

We thus have that $$ x^{50} = (x^2 - 2x + 1) P(x) + 50x - 49, $$ and so we have that $$ A^{50} = (A^2 - 2A + I) P(A) + 50A - 49I = 50A - 49I = \begin{pmatrix} 1 & 50 \\ 0 & 1 \end{pmatrix}. $$

Answer 2

You can compute $A^2,A^4,A^8,A^{16},A^{32}$ quite easily (repeated squaring).

Then compute $A^{50}=A^2A^{16}A^{32}$.

Answer 3

Hint

$$A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}+\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}=I_2+B$$ First matrix is identity matrice,the second one is nilpotent $B^2=0$

Answer 4

I guess there are several ways to proceed. One thing you could do is try to write down $A^2, A^3, A^4, \dots$ in terms of $A$ and $I$, spot a pattern, and use induction:

• $A^2 = 2A - I$
• $A^3 = 2A^2 - A = 2(2A - I) - A = 3A - 2I$
• $A^4 = 2A^3 - A^2 = 2(3A - 2I) - (2A - I) = 4A - 3I$

$\vdots$

Alternatively, you could repeatedly square:

• $A^2 = 2A - I$
• $A^4 = (A^2)^2 = (2A - I)^2 = 4A^2 - 4A + I = 4(2A - I) - 4A + I = 4A - 3I$
• $A^8 = (A^4)^2 = (4A-3I)^2 = \dots$

Answer 5

Note that $A$ is already in its Jordan form. Thus, for any differentiable function $f$ we have

$$f(A) = \begin{bmatrix} f(1) & f'(1) \\ 0 & f(1)\end{bmatrix}$$

In particular for $f(x) = x^{50}$ we have

$$A^{50} = \begin{bmatrix} 1 & 50 \\ 0 & 1\end{bmatrix}$$

Answer 6

The characteristic polynomial of $A$ is $(x-1)^2$, hence by the Cayley-Hamilton theorem $A^{n+2} = 2 A^{n+1}-A^{n}$ holds for any $n\in\mathbb{N}$. With respect to the entries of $A^n$ this implies that $\{A^n_{11}\}_{n\geq 0},\{A^n_{12}\}_{n\geq 0},\{A^n_{21}\}_{n\geq 0},\{A^n_{22}\}_{n\geq 0}$ are sequences of values of linear polynomials. Since $$A^0=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix},\quad A^1=\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}, \quad A^2=\begin{pmatrix}1 & 2 \\ 0 & 1\end{pmatrix}$$ by interpolation we get $A^{n}_{11}=A^{n}_{22}=1$, $A^n_{21}=0$ and $A^n_{12}=n$ for any $n\geq 0$.