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

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.

• It should be $A^3-I_2=\begin{bmatrix}3&3\\-3&-3\end{bmatrix}=3\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$, as we are not dealing with determinants here. – trancelocation Jul 2 '18 at 11:02

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 $$A^3 = \begin{bmatrix}4&3\\-3&-2\end{bmatrix} \Rightarrow A^2+A = \begin{bmatrix}5&3\\-3&-1\end{bmatrix}$$

• Is this answer legit? – C. Cristi Jul 2 '18 at 9:38
• @C.Cristi: This depends on what you define to be "legit". Surely this solution has an "experimental" component. But this happens often while solving math problems :-) – trancelocation Jul 2 '18 at 9:48
• Basically, you just "guessed" it, you know this is how it "should be" but what makes you think that you answer is actually correct? What makes your answer correct? Does this work for any matrix A_{nxn} or just 2x2? Can this pattern of problems repeat? – C. Cristi Jul 2 '18 at 9:50
• Can you be more detalied on how you applied it "backwards"? – C. Cristi Jul 2 '18 at 10:01
• "backwards" means here, that multiplying $A$ by $A^2$ has risen the entries of the upper row of $A$ by $2$ and reduced the entries of the lower row by $2$. So, subtract $2$ from the entries in the upper row of $A^3$. Add $2$ to the entries of the lower row of $A^3$. I do not claim that this strategy is a generally valid one. But it works here as your $A$ has a very special structure. – trancelocation Jul 2 '18 at 10:07

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.

• Can you add more details to your answer? – C. Cristi Jul 2 '18 at 6:23
• Why $\det(A-xI_2)=0?$ No, I mean how do you calculate $\det(A-xI_2)?$ – C. Cristi Jul 2 '18 at 6:29
• I don't get how $(4 - x)(-2 -x) + 9$ is the characteristic polynomial of $A$; it looks like the characteristic polynomial of $A^3$. – Robert Lewis Jul 2 '18 at 6:39
• Assuming that $A$ is a real matrix, you can easily see that $A$ has two eigenvalues $1$ and $1$ (repeated), and the rest follows similarly. So, you don't need to delete your answer. You have gone far, so it's better that you simply fix your answer a bit. If you want to fix it, I don't have to spend time writing my solution. – Batominovski Jul 2 '18 at 7:00
• Note that $A^3$ has a degenerate eigenspace associated to its unique eigenvalue $1$, so $A$ must also have a degenerate eigenspace associated to a cubic root of $1$ as the corresponding eigenvalue. If $A$ is real, then this cubic root of $1$ must be $1$ itself. – Batominovski Jul 2 '18 at 7:13

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.

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 $\,\,B_1 + B_2 + B_3 = 0.$