Let $A = \begin{pmatrix}0 & a\\ b&c\\ \end{pmatrix}$ with $a, b, c \in \mathbb{R}$. My question is that: ``How can we compute $A^n$ for any $n\in \mathbb{N}$?

In fact, one had that $A^2 - cA - abI_2 = 0$ or $A^2 = cA + abI_2$. Then I obtained \begin{eqnarray} A^2 &=& cA + abI_2\\ A^3 &=& (c^2 +ab)A + abcI_2\\ .....&&....................\\ \end{eqnarray}

  • 2
    $\begingroup$ Hint: Eigenvalue decomposition $\endgroup$ Feb 8, 2017 at 15:51
  • 3
    $\begingroup$ @mathJuan: Hint to the first hint diagonalize. $\endgroup$
    – Moo
    Feb 8, 2017 at 15:55
  • $\begingroup$ Diagonalise and then you can raise to any power by raising the diagonal matrix because the basis change matrices cancel out. $\endgroup$ Feb 8, 2017 at 16:00
  • $\begingroup$ Well, diagonal form or Jordan normal form. You'll need to work through the various cases if no constraints are given on the coefficients. It's not an enormous job, but probably too big for a proper answer here. And very, very standard. $\endgroup$ Feb 8, 2017 at 16:18
  • $\begingroup$ @Moo: Yes, I knew that but I hope someone can give a form of $A^n$ as a linear combination of $A$ and $I_2$. Because the diagonalise is many cases. $\endgroup$ Feb 8, 2017 at 18:46

2 Answers 2


The characteristic polynomial of this matrix is $\lambda^2-\lambda c-ab$, so you know that $A^2=cA+abI$. From there, you can use the Binomial Theorem to express $A^n$ as a linear combination of $A$ and $I$.

Update: This involves repeated expansion of terms of the form $(cA+abI)^k$, which can get pretty tedious for even moderately-sized $k$. Fortunately, for $2\times2$ matrices, there are only three (really four) cases to consider. Note that this analysis holds for $2\times2$ matrices in general, not just those that have the specific form in the question.

Case 1: $c^2+4ab\gt0$ (real distinct roots $\lambda_1$ and $\lambda_2$). The matrix can be diagonalized as $A=B\Lambda B^{-1}$, where $\Lambda=\operatorname{diag}(\lambda_1,\lambda_2)$, and so $A^n=B\Lambda^nB^{-1}=B\operatorname{diag}(\lambda_1^n,\lambda_2^n)\,B^{-1}$. In practice, there’s no need to diagonalize the matrix, however. Define $$P_1={A-\lambda_2I\over\lambda_1-\lambda_2}, P_2={A-\lambda_1I\over\lambda_2-\lambda_1}$$ so that $A=\lambda_1P_1+\lambda_2P_2$. It’s not hard to show that these two matrices are idempotent (in fact, they are projections onto the respective eigenspaces), and that they are complementary, in that $P_1P_2=P_2P_1=0$, so $$A^n = (\lambda_1P_1+\lambda_2P_2)^n = \lambda_1^nP_1^n+\lambda_2^nP_2^n = \lambda_1^nP_1+\lambda_2^nP_2$$ since all terms that involve both $P_1$ and $P_2$ vanish.

Case 2: $c^2+4ab=0$ (repeated real root $\lambda$). This is actually two cases. If $A=\lambda I$, then $A^n=\lambda^nI$. Otherwise, the matrix will have Jordan normal form $\pmatrix{\lambda&1\\0&\lambda}$. This can be written as $\lambda I+\pmatrix{0&1\\0&0}$. The second matrix is nilpotent of degree 2, so $$\pmatrix{\lambda&1\\0&\lambda}^n=\left(\lambda I+\pmatrix{0&1\\0&0}\right)^n=\lambda^nI+n\lambda^{n-1}I\pmatrix{0&1\\0&0}=\pmatrix{\lambda^n&n\lambda^{n-1}\\0&\lambda^n}$$ since the other terms in the binomial expansion vanish. So if $A=B\pmatrix{\lambda&1\\0&\lambda}B^{-1}$ then $A^n=B\pmatrix{\lambda^n&n\lambda^{n-1}\\0&\lambda^n}B^{-1}$. As in case 1, however, there’s no need in practice to compute a Jordan basis $B$ for the matrix, because just as we decomposed the Jordan matrix above, we can write $A$ as the sum of a multiple of the identity and a nilpotent matrix. Set $N=(A-\lambda I)$. By the Cayley-Hamilton Theorem, $N^2=(A-\lambda I)^2=0$, so $$A^n=(\lambda I-N)^n=\lambda^nI+n\lambda^{n-1}N$$ because terms involving $N^2$ or higher vanish.

Case 3: $c^2+4ab\lt0$ (complex roots). Since the characteristic polynomial has real coefficients, the two eigenvalues are of the form $\alpha\pm i\beta$. $A$ is then similar to the conformal matrix $C=\pmatrix{\alpha&-\beta\\\beta&\alpha}$, i.e., $A=BCB^{-1}$ and so $A^n=BC^nB^{-1}$. The matrix $C$ is isomorphic to the complex number $\alpha+i\beta$, so $C^n$ can be computed via DeMoivre’s theorem, or by writing $C=\alpha I+\beta J$, where $J^2=-I$, expanding via the binomial theorem and collecting terms.

As with the other two cases, there’s no need in practice to compute the above decomposition of $A$. We can instead write $A$ as the sum of a multiple of the identity and another matrix with nice properties. Let $G=A-\alpha I=\beta H$, where $H^2=-I$. (By the Cayley-Hamilton theorem, $(A-\alpha I)^2+\beta^2I=0$, so $G^2=-\beta^2I$). Then $$\begin{align}A^n&=(\alpha I+\beta H)^n=\binom n0\alpha^nI+\binom n1\alpha^{n-1}\beta H+\binom n2\alpha^{n-2}\beta^2H^2+\cdots \\ &=\left(\binom n0\alpha^n-\binom n2\alpha^{n-2}\beta^2+\cdots\right)I+\left(\binom n1\alpha^{n-1}\beta-\binom n3\alpha^{n-3}\beta^3+\cdots\right)H.\end{align}$$


For small matrices like that you can generally do it directly by identification.

Let set $A_n=\pmatrix{\alpha_n & \gamma_n\\\beta_n & \delta_n\\}$ with $\alpha_0=\delta_0=1$ and $\beta_0=\gamma_0=0$ the identity matrix $I=A^0$.

$A_{n+1}=A_nA=\pmatrix{\alpha_n & \gamma_n\\\beta_n & \delta_n\\}\pmatrix{0 & a\\b & c\\}=\pmatrix{b\gamma_n & a\alpha_n+c\gamma_n\\b\delta_n & a\beta_n+c\delta_n\\}$

We get the system of coefficients by identification $\begin{cases} \alpha_{n+1}=b\gamma_n \\ \beta_{n+1}=b\delta_n \\ \gamma_{n+1}=a\alpha_n+c\gamma_n \\ \delta_{n+1}=a\beta_n+c\delta_n \end{cases}$

In particular we can solve directly $\begin{cases} \gamma_{n+1}=ab\gamma_{n-1}+c\gamma_n\qquad\gamma_0=0,\gamma_1=a \\ \delta_{n+1}=ab\delta_{n-1}+c\delta_n\qquad\delta_0=1,\delta_1=c \end{cases}$

They both have the same recurrence characterictic equation $x^2-cx-ab=0$

Note: no surprise here, this is also $\det(A-xI)$ and we will use eigenvalues.

Let's have $\lambda_1,\lambda_2$ the eigenvalues (i.e. the roots of this quadratic equation).

We know that $\gamma_n$ and $\delta_n$ are of the form $u{\lambda_1}^n+v{\lambda_2}^n$, I pass the detail of calcultating $u,v$ given the initial conditions and go directly to the result (hoping I made no error...).

$\begin{cases} \alpha_{n}=b\gamma_{n-1} \\ \beta_{n}=b\delta_{n-1} \\ \gamma_{n}=\frac{a}{\lambda_1-\lambda_2}({\lambda_1}^n-{\lambda_2}^n) \\ \delta_{n}=\frac{c-\lambda_2}{\lambda_1-\lambda_2}{\lambda_1}^n+\frac{\lambda_1-c}{\lambda_1-\lambda_2}{\lambda_2}^n \end{cases}$


And then you pray for the eigenvalues to be real, else this is getting a bit annoying for practical calculus... They would be conjugated, though, if complex. You can alternately search for a form $\rho(u\cos(n\theta)+v\sin(n\theta))$

Also I haven't treated the case of a double root, in that case search for a form $(u+vn){\lambda}^n$, the rest of the process is identical.


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