Solving a simple matrix polynomial [closed]

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Does there exist a $2\times 2$ Matrix $A$ such that

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

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6 Answers

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Let $A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$, then $A^2 = \begin{bmatrix} a & b\\ c & d\end{bmatrix}\cdot \begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd\\ ac + cd & bc + d^2\end{bmatrix}$. Thus:

$A - A^2 = \begin{bmatrix} a-a^2-bc & b-ab-bd\\ c-ac-cd & d-bc-d^2\end{bmatrix} = \begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$. Thus:

$a-a^2-bc = 3$

$d-bc-d^2 = 4$

$b-ab-bd = 1$

$c-ac-cd = 1$.

Subtract the last $2$ equations to each other:

$b-c - a(b-c) - d(b-c) = 0 \to (b-c)(1-a-d) = 0$.

And subtract the first $2$ equations to obtain:

$a-d -(a^2-d^2) = -1 \to (a-d)(1-a-d) = -1$. Thus: $b-c = 0$, and $b = c$. Let's rewrite the system:

$a-a^2-b^2 = 3$

$d-b^2-d^2 = 4$

$b-ab-bd = 1$. Using Wolframalpha:

$a \approx 0.5 - 0.479i$

$b \approx 1.588i$

$d \approx 0.5 + 1.109i$.

Thus all roots are imaginary numbers, which means there is no matrix $A$ with real entries for the equation.

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The characteristic equation is $$(3-\lambda)(4-\lambda)-1=\lambda^2-7\lambda +11=0$$ has roots $$\frac{7\pm\sqrt{5}}{2}$$ which are both positive - so diagonalize the matrix as $A=S\Lambda S^{-1}$, and use the matrix $$S\Lambda^{1/2} S^{-1}$$

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Yes, there are complex solutions, but no real solutions. Using the brute force method also mentioned by JimmyK, you can see that $A$ must be symmetric. The top left element then becomes $a-a^2-b^2$ which must equal $3$, which can obviously only happen if $A$ has complex elements.

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This is a very unusual question. Let me try to give a conceptual proof that it has complex but no real solutions. First working over the complex numbers note that $\begin{pmatrix}3&1\\1&4\end{pmatrix}$ is diagonalizable as shown in NotNotLogical's answer (the eigenvalues are real). So conjugate the equation by $P$ and $P^{-1}$ so that it becomes, $$B-B^2=\begin{pmatrix}\lambda &0\\0&\mu\end{pmatrix}$$ where $B=PAP^{-1}$ and I write $\lambda$, $\mu$ for the eigenvalues. Now one can find diagonal $B$ by solving two quadratic equations over $\mathbb{C}$, $$x-x^2=\lambda$$ and $$y-y^2=\mu$$ Thus $A$ exists.

In any case $A$ must have a complex eigenvalue, say $\alpha$ it follows then that $\alpha-\alpha^2$ is an eigenvalue of $\begin{pmatrix}3&1\\1&4\end{pmatrix}$ from which it follows that $\alpha$ is not real. Therefore over $\mathbb{C}$ there are two distinct eigenvalues and thus $A$ is diagonalizable over $\mathbb{C}$. Now if $B$ is the diagonalization of $A$ over the complex numbers then $B-B^2$ is diagonal so by the uniqueness of diagonalization $B$ must satisfy the equation above. And this implies that the given matrix $\begin{pmatrix}3&1\\1&4\end{pmatrix}$ can be diagonalized by a complex matrix, but we also know that it is diagonalized by a real matrix, by uniqueness we have that $A$ is diagonalized by a real matrix and we know this is false, so there is no real solution.

Well its little messy, but maybe it can be streamlined.

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Completing the square, we find that the given equation is equivalent to $$ (A-\tfrac12 I)^2 = \tfrac14 I - \left[\begin{matrix} 3 & 1 \\ 1 & 4 \end{matrix}\right] = -\frac14 \left[\begin{matrix} 11 & 4 \\ 4 & 15 \end{matrix}\right] \tag{$\ast$} $$ If you wanted $A$ to be real, we can now show it can't be done by setting $$ A-\tfrac12 I = \left[\begin{matrix} a & b \\ c & d\end{matrix}\right] $$ which yields $$ \left[\begin{matrix} a^2+bc & b(a+d) \\ c(a+d) & bc+d^2 \end{matrix}\right] = \left[\begin{matrix} -11/4 & -1 \\ -1 & -15/4 \end{matrix}\right] $$ Since $b(a+d)$ and $c(a+d)$ are equal (and not zero), we get $b=c$; but then the diagonal entries are $a^2+b^2$ and $b^2+d^2$, which are nonnegative, contrary to what we see.

If you don't mind $A$ having complex entries, you can note that the RHS of ($\ast$) is symmetric, so it's diagonalizable; thus it has a (complex) square root, and adding $\frac12 I$ to that square root will give $A$.

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Assume such an $A$ exists. Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$. Then, the eigenvalues of $A-A^2$ are $\lambda_1-\lambda_1^2$ and $\lambda_2-\lambda_2^2$.

The eigenvalues of $\begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$ are $\dfrac{7 \pm \sqrt{5}}{2}$. Hence $\lambda_1-\lambda_1^2 = \dfrac{7 - \sqrt{5}}{2}$ and $\lambda_2-\lambda_2^2 = \dfrac{7 + \sqrt{5}}{2}$.

Solving gives us two possibilities for each eigenvalue:

$\lambda_1 = \dfrac{1}{2} \pm i\dfrac{1}{2}\sqrt{13-2\sqrt{5}}$ and $\lambda_2 = \dfrac{1}{2} \pm i\dfrac{1}{2}\sqrt{13+2\sqrt{5}}$.

If the entries of $A$ are all real, then either $\lambda_1, \lambda_2 \in \mathbb{R}$ or $\lambda_1 = \overline{\lambda_2}$, which is not the case here.

Hence, there is no real $2 \times 2$ matrix $A$ such that $A-A^2 = \begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$.

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