Compute $A^n$ where $A^2+bA+cI=0$

Question

Let $A$ be a complex matrix such that $$A^2+bA+cI=0,$$ where $I$ is the identity matrix and $b,c\in \mathbb{C}$. I am interested in finding a formula for $A^n$ in terms of $A$ and $I$. The binomial formula is not giving an answer I think. Maybe using $A^2=-bA-cI$, then $A^3=-bA^2-cA$, etc. But the computation becomes so complicated to find a formula by induction.

Answer 1

Hint By Long Division

$$X^n=(X^2+bX+c)Q(X)+\alpha X+ \beta$$

Plug in the roots of $X^2+bX+c=0$ to find $\alpha, \beta$.

Then plug in $A$.

P.S. If the two roots are equal, then we have $$X^n=(X-\gamma)^2Q(X)+\alpha X+\beta \tag{*}$$

After puggin in $\gamma$ in $(*)$ you need to differentiate $(*)$ and plug in $\gamma$ again.

This method works even if the polynomial you have has higher degree.

Answer 2

You have $$ A^{n+2} = b_nA + c_nI $$

where $b_0 = -b, c_0 = -c$

Multiplying both sides by $A$ gives $$ A^{n+3} = b_nA^2 + c_n A = b_n(-bA-cI)+c_nA = (c_n-bb_n)A - cb_nI $$

therefore $$ b_{n+1} = -bb_n + c_n, \ c_{n+1} = -cb_n $$

You may find an implicit relation by solving the linear system $$ \left(\matrix{b_n\\c_n}\right)=M^n\left(\matrix{-b\\-c}\right) $$

where $$ M = \left(\matrix{-b&c\\-c&0}\right) $$

then we have the characteristic $$ \lambda^2 + b\lambda + c^2 = 0 $$

The two eigenvalues are $\lambda_1,\lambda_2$, where $b = -(\lambda_1+\lambda_2), c^2 = \lambda_1\lambda_2$

The eigenvectors are $$ \vec{v_1} = \left(\matrix{-\lambda_1\\c}\right), \ \vec{v_2} = \left(\matrix{-\lambda_2\\c}\right) $$

Thus $$ M^n\left(\matrix{-b\\-c}\right) = -M^n(\vec{v_1}+\vec{v_2}) = -{\lambda_1}^n\ \vec{v_1} - {\lambda_2}^n\ \vec{v_2} $$

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If $\lambda_1 = \lambda_2$ then we have the generalized eigenvectors $$ \vec{v_1} = \lambda \left(\matrix{-1\\1}\right), \ \vec{v_2} = \left(\matrix{-2\\1}\right) $$

Then $$ M^n\left(\matrix{2\lambda\\-\lambda}\right) = -M^n \lambda \vec{v_2} = \lambda^{n+1}\vec{v_2} + n\lambda^n \vec{v_1} $$

Answer 3

The polynomial $m(\lambda)=\lambda^2+b\lambda +c$ annihilates $A$. Therefore, \begin{align} (A+(\mu+b)I)(A-\mu I)&=A^2+bA-\mu(\mu+b)I\\ &= -(\mu^2+b\mu+c)I \\ (A-\mu I)^{-1}&=-\frac{1}{\mu^2+b\mu +c}(A+(\mu+b)I). \end{align} Using a little Complex Analysis by integrating in $\mu$ around both zeros of $\mu^2+b\mu +c$ gives $$ f(A)=\frac{1}{2\pi i}\oint_{C}f(\mu)(\mu I-A)^{-1}d\mu. $$ The zeroes of $\mu^2+b\mu +c$ are $-b/2\pm \sqrt{b^2/4-c}$ denoting these roots by $r_1,r_2$ gives an integral $$ f(A) = \frac{1}{2\pi i}\oint_{C}f(\mu)\frac{1}{(\mu-r_1)(\mu-r_2)}(A+(\mu+b)I)d\mu \\ = f(r_1)\frac{1}{r_1-r_2}(A+(r_1+b)I)+f(r_2)\frac{1}{r_2-r_1}(A+(r_2+b)I) \\ = \frac{f(r_1)-f(r_2)}{r_1-r_2}(A+bI)+\frac{r_1f(r_1)-r_2f(r_2)}{r_1-r_2}I. $$ So, for example, $$ A^{n} = \frac{r_1^n-r_2^n}{r_1-r_2}(A+bI)+\frac{r_1^{n+1}-r_2^{n+2}}{r_1-r_2}I. $$