Show that there exists some $\epsilon >0$ s.t. for all $\mathbf{A}\in \mathbb{R}^{2\times 2} $ with $|( \mathbf{A})_{i, j}| < \epsilon $ for all $i, j$ (let the space of all such matrices be $E$) the equation $$ \begin{align*} \mathbf{X}^{2} + \mathbf{X} = \mathbf{A} \end{align*} $$ has some solution $\mathbf{X}\in \mathbb{R}^{2\times 2} $.

My approach would be to use the Banach fixed point theorem: Consider the mapping $$ \begin{align*} \Phi \colon E \to \mathbb{R}^{2, 2}, \quad \Phi ( \mathbf{X}) = \mathbf{A} - \mathbf{X}^{2} .\end{align*} $$ So first we need to find an $\epsilon > 0$ s.t. $\Phi [ E]\subseteq E$. Since $$ \begin{align*} \begin{bmatrix} a & b \\ c & d \end{bmatrix} ^{2} = \begin{bmatrix} a^{2} + cb & ab + db \\ ac + c d & b c + d ^{2} \end{bmatrix} \implies \forall i, j \in \{ 1, 2\}\colon ( \mathbf{X}^{2})_{i, j} < 2 \epsilon^{2} \end{align*} $$ choosing any $\epsilon < 1 / 2$ will assure that $\Phi ( \mathbf{X}^{2}) \in E$ for $\mathbf{X} \in E$. Next we have to determine when $\Phi $ becomes a contraction. One has $$ \begin{align*} \left\| \Phi ( \mathbf{X}) - \Phi ( \mathbf{Y})\right\|_{\infty } &= \left\| \mathbf{X}^{2} - \mathbf{Y}^{2}\right\|_{\infty} \\ &= \max_{i \in \{ 1, 2\}} \sum_{j = 1}^{2} \left| ( \mathbf{X}^{2})_{i, j}- ( \mathbf{Y}^{2})_{i, j} \right| \\ &\leqslant \max_{i \in \{ 1, 2\}} \sum_{j = 1}^{2}\!\left( \left| ( \mathbf{X}^{2})_{i, j}\right|+ \left| ( \mathbf{Y}^{2})_{i, j} \right| \right) \\ & < 2 \!\left( 2 \epsilon ^{2} + 2 \epsilon ^{2}\right) = 8 \epsilon ^{2} \end{align*} $$ and $\left\| \mathbf{X} - \mathbf{Y}\right\|_{\infty} < 4\epsilon $ meaning we simply have to choose some $\epsilon $ satisfying $$ \begin{align*} 2 \epsilon ^{2} < \epsilon \iff \epsilon < \frac{1}{2} .\end{align*} $$ First of all, is my approach correct? Secondly, are there more elegant solutions (this question was in the context of an analysis course, meaning I didn't think about any elegant linear algebra solutions).

  • $\begingroup$ Have you tried using the implicit function theorem? Clearly, (0,0) is a solution. Then, you should be able to find that there are also solutions in a neighborhood of that point. $\endgroup$
    – KBS
    Aug 8 at 15:49
  • $\begingroup$ @KBS I'll give this idea a try later! But does my approach also work? $\endgroup$
    – Richard
    Aug 8 at 15:56
  • $\begingroup$ So $(A)_{ij}$ may have arbitrary negative values? And you should mention that $E$ depends on $\epsilon$. $\endgroup$
    – Paul Frost
    Aug 8 at 16:18

3 Answers 3


Your approach is not correct, for the simple reason that you haven't showed $\Phi$ is a Banach contraction. You have provided upper bounds for both $\|X - Y\|_\infty$ and $\|\Phi(X) - \Phi(Y)\|_\infty$ in terms of $\varepsilon$, but you have not related them. Using triangle inequality to turn a difference to a sum is a big red flag in this regard!

As for an alternate, more linear algebra-flavoured approach, we can use the following results:

  • Suppose $\| \cdot \|$ is a matrix norm and $\|B - I\| < 1$. Then $B$ is invertible.
  • If $B$ is invertible, then $B$ has a complex square root.

The former is a classical result; the inverse of $B$ can be expressed as a geometric series of matrices: $$B^{-1} = \sum_{n=0}^\infty (I - B)^n.$$ It can also be proven using this result; because $\|B - I\| < 1$, we know the largest eigenvalue of $B - I$ is less than $1$. This means $1$ is not an eigenvalue, so $B = B - I + I$ is invertible!

The latter can be proven with Jordan normal forms. Each Jordan block takes the form $\lambda I + N$, where $\lambda$ is the eigenvalue for the Jordan block, and $N$ is a nilpotent matrix (i.e. the $1$s along the superdiagonal). One can then substitute $\frac{1}{\lambda}N$ into the Taylor series for $\sqrt{1 + x}$ (note: as $N$ is nilpotent, only finitely many terms are non-zero) to get a square root of $\frac{1}{\lambda}(\lambda I + N)$. Multiply this square root by $\sqrt{\lambda}$, and you get the square root of the Jordan block. Assembling these square roots in block-diagonal form gives a square root of the JNF of the matrix, which will be similar to a square root of the original matrix.

So, using these facts, we can show that, if $\varepsilon < \frac{1}{8}$, then $X^2 + X = A$ has a solution. To see this, complete the square: $$\left(X + \frac{1}{2}I\right)^2 = A + \frac{1}{4}I = \frac{1}{4}(4A + I).$$ If $|(A)_{ij}| < \varepsilon$, then $\|A\|_\infty < 2\varepsilon < \frac{1}{4}$. Thus, $\|4A\|_\infty < 1$, so by the first result, $4A + I$ must be invertible. By the second result, $4A + I$, and hence $A + \frac{1}{4}I$, must have a square root. Let $B$ be a square root of the latter. This gives us a solution: $$X = B - \frac{1}{2} I.$$

  • $\begingroup$ true, thanks! I think my mistake is not easy to rectify. $\endgroup$
    – Richard
    Aug 8 at 16:28
  • $\begingroup$ what do you think of my modified argument? $\endgroup$
    – Richard
    Aug 8 at 20:11
  • $\begingroup$ @Richard Matrix calculus is not one of my strong points, sorry. $\endgroup$ Aug 8 at 23:23

I have a short nonconstructive proof that works in $n$ dimensions.

Let $F:\mathcal{B}\times\mathcal{B}\mapsto\mathcal{B}$ be defined as $F(A,X) = X^2+X-A$ where $\mathcal{B}$ is the Banach space of matrices in $\mathbb{R}^{n\times n}$ with any matrix norm $\|\cdot\|$ (which are all equivalent as we are in finite dimensions); e.g. the induced $\infty$-norm.

Note further that $F(0,0) = 0$, so $(0,0)$ is obviously a solution to the equation. Now, this map is continuously Fréchet differentiable and its Fréchet derivative at $(A,X)$ is given by


where $H_1,H_2\in\mathbb{R}^{n\times n}$ and we have that the map $H_2\mapsto L(0,0)[0,H_2]=H_2$ is an isomorphism from $\mathcal{B}$ to itself as it is nothing else but the identity map.

Therefore, by virtue of the implicit function theorem, there exist neighborhoods $U\subset\mathbb{R}^{n\times n}$ of $0$ and $V\subset\mathbb{R}^{n\times n}$ of $0$, and a Fréchet differentiable function $G : U\mapsto V$ such that $F(A, G(A)) = 0$ and $F(A, X) = 0$ if and only if $X = G(A)$, for all $\displaystyle (A,X)\in U\times V$.

Then, the conclusion follows that one can find a small enough $\epsilon>0$, such that

$$\{A\in\mathbb{R}^{n\times n}: |A_{ij}|<\epsilon\}\subset U.$$

In some sense, this result is expected since in a sufficiently small neighborhood of the origin, the equation behaves like a linear equation and so the local existence of solutions is guaranteed. This easily generalizes to more complex expressions of the form


where $m\in\mathbb{N}$, $\alpha_i\in\mathbb{R}$, $i=1,\ldots,m$ and $\alpha_1\ne0$.

  • $\begingroup$ +1 that's what I came up with in the $2 \times 2$ case above after reading a comment proposing the implicit function theorem. $\endgroup$
    – Richard
    Aug 8 at 19:35
  • $\begingroup$ @Richard That was my comment which I decided to turn into an answer :) $\endgroup$
    – KBS
    Aug 8 at 19:48
  • $\begingroup$ I know, thanks for the comment! Yes, in my above comment it should've been "the comment" instead of "a comment". $\endgroup$
    – Richard
    Aug 8 at 19:52

Lemma: There exists $\epsilon>0$ such that if $B$ is a $2\times 2$ matrix with $\|B-\frac{1}{4}I \|_\infty<\epsilon$, then the equation $Y^2=B$ has a solution $Y$.

Proof: Since the eigenvalues of $S$ are given by $$\lambda_{1,2}(S)=\frac{1}{2} \left(S_{11}+S_{22}\pm \sqrt{(S_{11}+S_{22})^2-4(S_{11}S_{22}-S_{12}S_{21})} \right ),$$ and $\frac{1}{4}I$ has positive eigenvalues, there exists a neighborhood of $\frac{1}{4}I$ such that every matrix in that neighborhood whose eigenvalues are real must have positive eigenvalues as well. In other words, there exists $\epsilon>0$ such that for every matrix $B$ with $\|B-\frac{1}{4}I\|_\infty<\epsilon$, either $B$ has non-real eigenvalues or $B$ has real positive eigenvalues. Next, by Jordan decomposition, there exists a matrix $P$ such that $P^{-1}BP=\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}$ with $\lambda_1,\lambda_2>0$ (if $B$ has two positive eigenvalues) or $P^{-1}BP=t\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$ for some $t>0$ and angle $\alpha$ (if $B$ has non-real eigenvalues). In either case, it is easy to see that there exists $Z$ with $Z^2=P^{-1}BP$. To be precise, in the former case, let $Z=\begin{bmatrix} \sqrt{\lambda_1} & 0 \\ 0 & \sqrt{\lambda_2}\end{bmatrix}$ and in the latter case, let $Z=\sqrt{t}\begin{bmatrix} \cos (\alpha/2) & -\sin (\alpha/2) \\ \sin (\alpha/2) & \cos (\alpha/2) \end{bmatrix}$. We let $Y=PZP^{-1}$ to get $Y^2=B$.

Using this lemma, now let $B=A+\frac{1}{4}I$. Hence there exists $Y$ with $Y^2=B$. Finally, let $X=Y-\frac{1}{2}I$ to get $X^2+X=A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.