x$A^{100 }$ where $A = \begin{bmatrix} 1 &2 \\ 3& 4 \end{bmatrix}$

Question

Compute $A^{100 }$ where $A = \begin{bmatrix} 1 &2 \\ 3& 4 \end{bmatrix}$.

I can calculate $A^{100}$ using a calculator, but my question is that is there any short formula/method or is their any trick to find the $A^{100}$?

Answer 1

You could use diagonalization.

Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PBP^{-1}$ with $B$ as a diagonal matrix and $P^{-1}$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PB^2P^{-1}$. Similarly for $A^n=PB^nP^{-1}$

$A^{100}=PB^{100}P^{-1}$

Answer 2

You can first diagonalize the matrix as follows: $$A=P^{-1}DP,$$ where $P$ is an orthogonal matrix. The matrix $P$ is the matrix of eigenvectors $\{v_1,v_2\}$ that correspond to eigenvalues $\{\lambda_1,\lambda_2\}$ of $A$. Here, $$D=\mathrm{diag}\{\lambda_1,\lambda_2 \}$$

After diagonalizing, you can calculate $A^{100}$ as follows: $$A^{100} = (P^{-1}DP)^{100}=P^{-1}D^{100}P = P^{-1}\mathrm{diag}\{\lambda_1^{100},\lambda_2^{100}\}P. $$

Answer 3

The conventional answer is going to be to diagonalize the matrix into $A=P\Lambda P^{-1}$ and then compute $P\Lambda^{100}P^{-1}$, but once you have the eigenvalues $\lambda_1$ and $\lambda_2$ there are ways to do this without computing an eigenbasis:

• Decompose $A$ into $\lambda_1P_1+\lambda_2P_2$, where $P_1$ and $P_2$ are projections onto the corresponding eigenspaces with $P_1P_2=P_2P_1=0$. There’s a fairly simply formula for these projections in terms of $A$ and the two eigenvalues. If you expand $A^{100}$ using the binomial theorem, you’ll find that all but two terms vanish.
• Use the Cayley-Hamilton theorem to write $A^{100}=aI+bA$ for some undetermined coefficients $a$ and $b$. This equation is also satisfied by the eigenvalues, which gives you the system of linear equations $a+b\lambda_i=\lambda_i^{100}$ for $a$ and $b$.

The above assumes that $A$ has distinct real eigenvalues. If they’re not, you will have to modify the above methods a bit.

Answer 4

There is no need for diagonalizations.

The eigenvalues of $$A = \begin{bmatrix} 1 &2 \\ 3& 4 \end{bmatrix}$$ are $$\frac {5\pm\sqrt {33}}{2}$$

Cayley-Hamilton Theorem indicates that $$A^{100}=\alpha A + \beta I.$$

We can find the coefficients $\alpha$ and $\beta $ by equations

$$ \alpha \lambda _1 +\beta = \lambda _1^{100}\\ \alpha \lambda _2 +\beta = \lambda _2^{100}$$ Where $\lambda _1$ and $\lambda _2$ are eighenvalues of $A.$

Answer 5

Hint: the characteristic function of the matrix is $$\lambda^2=5\lambda+2$$so according to Caylay-Hamilton theorem we have $$A^2=5A+2$$