Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$

Question

Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$

Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt:

$$A^3 -I_2 = \begin{bmatrix} 3 & 3\\ -3 & -3\end{bmatrix} = 9 \begin{bmatrix} 1 & 1\\ -1 & -1\end{bmatrix}$$

and

$$B=\frac {A^3-I_2}{A-I_2}-I_2$$

$$(A-I_2)(A^2+A+I_2)=9\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$$

But I get stuck here.

Answer 1

Although I prefer an algebraic approach, here is a "non-algebraic" way by directly determining $A$. Let

• $C =A^3 \Rightarrow C^2=\begin{bmatrix}7& 6\\-6&-5\end{bmatrix}$

Looking at the pattern of the entries comparing $A^6$ with $A^3$ we find:

• the entries in the first row go $3$ up
• the entries in the second row go $3$ down

So, applying this pattern "backwards" to $A^3$ we get a possible candidate for $A$: $$A = \begin{bmatrix}2& 1\\-1&0\end{bmatrix}$$ Indeed, we get $$A^3 = \begin{bmatrix}4&3\\-3&-2\end{bmatrix} \Rightarrow A^2+A = \begin{bmatrix}5&3\\-3&-1\end{bmatrix}$$

Answer 2

The matrix $C=A^3$ satisfies its characteristic equation, that is: $det(A^3-xI)=0$, which is $(4-x)(-2-x)+9=0$. See https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem.

This gives $C^2-2C+I=0$, thus $C$ has the eigenvalue 1. The eigenvalues of $A^3$ are the cubes of the eigenvalues of $A$. If the eigenvalues of $A$ are the two complex cube roots of unity, then we'd get $A^2+A=-I$, but then $A^3-I$ is non-zero, thus the eigenvalues of $A$ are also both 1, which means that $A$ satisfies $A^2=2A-I$.

This gives $A^3=2A^2+A=2(2A-I)+A=3A-2I$. Thus, $A^2+A=3A-I=A^3+I$.

Remark: The conclusion that the eigenvalues of $A$ must both be 1 is wrong, as they could be any of the other two cube roots of unity, occuring twice. This is shown in the other answers.

Answer 3

Presumably $A$ is real. Observe that $A^3-I$ is nonzero but nilpotent (because $(A^3-I)^2=0$). Therefore $A^3$ and in turn $A$ are not diagonalisable. Hence $A$ has repeated eigenvalues. So, if $A$ is real, its eigenvalues must be real (or else the trace of $A$ would become non-real) and equal to $1$ (because $A^3-I$ is nilpotent). Hence the Jordan form of $A$ is $\pmatrix{1&1\\0&1}$. Since $$ \pmatrix{1&1\\0&1}^2+\pmatrix{1&1\\0&1}=\pmatrix{2&3\\0&2} =\pmatrix{1&1\\0&1}^3+I, $$ we conclude that $A^2+A=A^3+I$.

Remark. Note that the above conclusion does not hold when $A$ can be non-real. (Thus the other answers here are either wrong or incomplete.) E.g. suppose $w=\exp(2\pi i/3)$ and $$ A=\pmatrix{2w&w\\ -w&0}. $$ Then $A^3$ is indeed equal to $\pmatrix{4&3\\ -3&-2}$ but $(A^2+A)-(A^3+I)$ is non-real.

Answer 4

Denote the third root of unity as $w=e^{i2\pi/3}$.

Then any of the quantities $\rho=\{w,w^2,w^3=1\}\,$ satisfy $\,\rho^3=1,\,$ i.e. is a cube root of one.

Note that the matrix $X=A^3$ has a single eigenvalue of $\{\lambda=1\}$ with multiplicity two.

Since it is a $2\times 2$ matrix, any analytic function of $X$ (and its first derivative) can be written as linear polynomials $$\eqalign{ f(X) &= \alpha_1X + \alpha_0I \cr f'(X) &= \alpha_1I \cr }$$ Evaluate the cube root function on the eigenvalues to calculate the polynomial coefficients $$\eqalign{ \alpha_1\lambda + \alpha_0 &= \lambda^{1/3} &\implies \alpha_1+\alpha_0 &= \rho \cr \alpha_1 &= \tfrac{1}{3}\lambda^{-2/3} &\implies \alpha_1 &= \tfrac{\rho}{3} \cr }$$ The matrix cube root is therefore $$\eqalign{ A &= f(X) = \,\,\frac{\rho}{3}X + \frac{2\rho}{3}I \cr }$$ and the matrix in question is $$\eqalign{ B &= A^2 + A \cr &= (\tfrac{\rho}{3}X + \tfrac{2\rho}{3}I)^2 + (\tfrac{\rho}{3}X + \tfrac{2\rho}{3}I) \cr &= \tfrac{\rho^2}{9}(X^2+4X+4I) + \tfrac{\rho}{3}(X+2I) \cr &= \frac{1}{9}\Big(\rho^2X^2+(4\rho^2+3\rho)X+(4\rho^2+6\rho)I\Big) \cr\cr }$$ This gives us 3 possible solutions for the $B$ matrix:

Setting $\rho=w^1=w\,$ yields $$B_1=\frac{1}{9}\Big(w^2X^2+(w^2-3)X+(2w-4)I\Big)$$ Setting $\rho=w^2\,$ yields $$B_2=\frac{1}{9}\Big(wX^2+(w-3)X+(2w^2-4)I\Big)$$ Setting $\rho=w^3=1\,$ yields $$B_3=\frac{1}{9}\Big(X^2+7X+10I\Big)$$

Some properties, peculiar to roots-of-unity, were used to simplify the final expressions. Namely $$\eqalign{ w &= w^4 \cr 0 &= 1+w+w^2 \cr 0 &= 1+w^2+w^4 \cr\cr }$$ Interestingly, these properties also mean that $\,\,B_1 + B_2 + B_3 = 0.$