$X^2 + X =A$ with $X, A\in \text{Mat}_{2,2} (\mathbb{R})$ . Show that there exists a solution $X$ for a given $A$

Question

Prove the following statement:

There exists an $\epsilon > 0$ such that the following holds:

If $A = (a_{ij}) \in \text{Mat}_{2,2} (\mathbb{R})$ a matrix with $(|a_{ij}|) < \epsilon$ for $i,j \in \{1,2\}$, then the following equation

$$X^2 + X = A$$

has a solution $X \in \text{Mat}_{2,2} (\mathbb{R})$

My Idea on how to solve this:

Let $X = \begin{bmatrix} v& w \\ x & y \end{bmatrix}$. Therefore $X^2 + X = \begin{bmatrix} v^2 + v + w x& v w + w y + w\\ v x + x y + x & w x + y^2 + y \end{bmatrix} = \begin{bmatrix} a_0 & a_1 \\ a_2 & a_3\end{bmatrix}$

Lets now define the function $$ 0=h(v,w,y,x,a_0,a_1,a_2,a_3) =\begin{cases} v^2 + v + w x - a_0 \\ v w + w y + w - a_1\\v x + x y + x -a_2 \\w x + y^2 + y-a_3 \end{cases} $$

We can now calculate the derivative of $h$:

$$dh = \begin{bmatrix} 2v + 1 & x & 0 & w & -1&0&0&0\\ w& v+y+1& w& 0& 0&-1&0&0\\ x & 0&x&v +1 & 0&0&-1&0 \\0&x&2y+1&w& 0&0&0&-1 \end{bmatrix}$$

The idea now would be to apply the implicit function theorem and show that there exists an $X$ which solves this equation. I am not sure though if this approach is correct.

Last but not least.. this question comes from an analysis sheet, so I assume one should use the methods of analysis to solve it.

Is my approach the correct way? And how does one proceed from here?

Feel free to use another approach.

Thank you for your time.

Answer 1

Identify the matrix $X=\begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix}$ with the vector $\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}.$

The the mapping $X^2 + X$ is equivalent to the following mapping from $\mathbb{R}^4$ to $\mathbb{R}^4$

$$ x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \to f(x) = \begin{pmatrix} x_1^2 + x_2x_3+x_1\\ x_1x_2+x_2x_4+x_2\\ x_3x_1 + x_3x_4+x_3\\ x_3x_2+x_4^2 +x_4\\ \end{pmatrix} . $$

The Jacobian matrix $f'(x)$ is a continuous function of $x$ and is given by $$ \begin{pmatrix} 2x_1+1 & x_3 & x_2 & 0\\ x_2 & x_1+x_4+1& 0 & x_2\\ x_3 & 0 & x_1 + x_4 + 1 & x_3 \\ 0 & x_3 & x_2 & 2x_4 + 1\\ \end{pmatrix}. $$

If $\mathbf{0}$ denotes the zero vector, then it is easy to see $f'(\mathbf{0})$ is the identity matrix, and by continuity it follows that there is a open set, $U$ containing the origin on which $f'(x)$ is invertible, and hence it follows (Apostal Mathematical analysis, 13.5 for example) that $f$ is an open mapping on $U$, and in particular the image of $U$ contains a sufficiently small open ball around $f(0) = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$

So $[ -\epsilon, \epsilon]^4 $ lies in the image of $U$ for sufficiently small $\epsilon > 0$ and the result follows.

Answer 2

This is a sketch about how to do it from a "pure linear algebra approach".

Case 1: suppose that $A$ have two real eigenvalues, then exists a change of basis $B\in \Bbb R^{2\times 2}$ such that $A_B:=BAB^{-1}$ is upper triangular.

Then it is enough to show that exists an upper triangular $X_B\in \Bbb R^{2\times 2}$ such that

$$X^2_B+X_B=X_B(X_B+I)=A_B\tag1$$

for suitable $\epsilon>0$.

Case 2: suppose that $A$ have two non-real eigenvalues, then exists a change of basis $B\in \Bbb R^{2\times 2}$ such that $A_B$ have the form $\left[\begin{smallmatrix}a&-b\\b&a\end{smallmatrix}\right]$. Then it is enough to show that exists some $X_B\in \Bbb R^{2\times 2}$ with the same form such that $(1)$ holds for some suitable $\epsilon>0$.

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The calculations, of the linear equations defined after these change of basis, are very easy to handle and the proof is straightforward.

Answer 3

My simple-minded way.

If $X^2 + X = A$ then $X^2 + X+\frac14 I = A+\frac14 I$ so $(X+\frac12 I)^2 =A+\frac14 I$.

Then use the method referenced by Dietrich Burde to find $\sqrt{A+\frac14 I}$ and then get $X$.

Answer 4

Alternatively (but essentially equivalent) to the other analytical-oriented answer, you can instead compute the derivatives more intrinsically: Consider the map $$f:M_n \times M_n \to M_n$$ $$(A,X) \mapsto X^2+X-A.$$ The derivative $D_{2,(A,X)}f$ (i.e., with respect to the second coordinate*) is given by $$H \mapsto XH+HX+H.$$ If $(A,X)=(0,0)$, then this is simply $H \mapsto H$, an isomorphism (the identity! Note the similarity with the other answer I mentioned), and the result then follows from the implicit function theorem. (Trivial, but must be mentioned: note that $f(0,0)=0$).

Arguably simpler is the approach mentioned in the comments: just consider $$f: M_n \to M_n$$ $$X \mapsto X^2+X.$$ The derivative of this map is $H \mapsto XH+HX+H$, which at zero is $H \mapsto H$, an isomorphism (again, the identity), and then the result follows from the inverse function theorem. Again, it must be mentioned that $f(0)=0$.

"Arguably" simpler because those approaches are morally and practically equivalent.

*If this is not clear, to be explicit: $D_{2,(A,X)}f$ is the derivative of the map $Y \mapsto f(A,Y)$ at $X$.

Answer 5

My way of attacking this problem uses both some algebra, and some analysis; also, there is nothing in the following which requires we restrict ourselves to $M_2(\Bbb R)$, the set of $2 \times 2$ real matrices; everything we say applies equally to $M_n(\Bbb R)$.

First, some algebra: the equation

$X^2 + X = A \tag 1$

has a solution $X$ for some $A$ if and only if the equation

$X^2 + X + \dfrac{1}{4} I = \dfrac{1}{4}I + A \tag 2$

has the same solution $X$ as well; also, since

$(X + \dfrac{1}{2}I)^2 = X^2 + X + \dfrac{1}{4}I, \tag 3$

we may write (2) as

$(X + \dfrac{1}{2}I)^2 = \dfrac{1}{4} I + A, \tag 4$

which has a solution precisely when there exists a matrix $C$ such that

$C^2 = \dfrac{1}{4}I + A, \tag 5$

in which event we may take

$X + \dfrac{1}{2} I = C, \tag 6$

or

$X = C - \dfrac{1}{2}I. \tag 7$

The reader informed in the machinations of high-school algebra will recognize the above process as completing the square of the polynomial $X^2 + X$.

The preceding discussion indicates that we need to resolve the question of the existence of $C$ satisfying (5) when the entries of $A$ are sufficiently small. To this end we turn to the tools of analysis. Consider the mapping

$B \mapsto B^2 \tag 8$

defined in some neighborhood of $\frac{1}{2}I$; for any linear map $H$ we have

$(B + H)^2 = B^2 + HB + BH + H^2, \tag 9$

whence

$(B + H)^2 - B^2 - (HB + BH) = H^2, \tag{10}$

whence

$\Vert (B + H)^2 - B^2 - (HB + BH) \Vert = \Vert H^2 \Vert \le \Vert H \Vert^2; \tag{11}$

since the right-hand side of (11) is of the from $\Vert H \Vert o(\Vert H \Vert)$, we find that the derivative $DB^2$ at any $B$ is the linear map

$DB^2(H) = HB + BH; \tag{12}$

with

$B = \dfrac{1}{2}I, \tag{13}$

(12) becomes

$DB^2(H) = H \left (\dfrac{1}{2}I \right ) + \left (\dfrac{1}{2}I \right ) H = H, \tag{14}$

that is, $DB^2$ is the identity mapping when (13) binds. We may now invoke that powerful tool of analysis, the inverse function theorem, to infer that there is a neighborhood $U$ of $\frac{1}{4}I$ and a continuously differentiable function

$S:U \to M_n(\Bbb R), \; S \left ( \dfrac{1}{4} \right ) = \dfrac{1}{2}, \tag{15}$

satisfying, for every $E \in U$,

$(S(E))^2 = E; \tag{16}$

that is, $S$ is an effective square-root on $U$.

The hypothesis placed upon $A$, that $\vert a_{ij} \vert < \epsilon$ for all $i, j$, allows us to conclude that for $\epsilon$ sufficiently small

$\dfrac{1}{4} I + A \in U; \tag{17}$

thus we may take

$C = S \left (\dfrac{1}{4}I + A \right ); \tag{18}$

then

$C^2 = \dfrac{1}{4}I + A, \tag{19}$

and with $X$ as in (7) we have a solution to (1).