# $A^2X = X$ show that $AX = X$, if $A_{ij} > 0$ and $x_{i} > 0$

Let $A$ and $X$ be matrices $n \times n$ and $n \times 1$, respectively, with all entries real and strictly positive. Assume that $A^2 X = X$. Show that $A X = X$.

What I thought:

(i) $A^2 X = X \Rightarrow A (A^2 X) = A (X)$ for distributive law $A(A^2 X) = A A (Ax) = AX \Rightarrow A^2(AX) = (AX)$, i.e., $A^2 Y = Y$ for $Y=AX$.

(ii) $A^2 X = X \Rightarrow (A^2-I)X = 0$. Be $X = (x_1,x_2,\ldots,x_n)^t$. So, $x_i > 0 \quad\forall i \in \{1,2,\ldots,n\}$. So, $det(A^2 - I) = 0 \Rightarrow det(A-I) = 0$ or $det(A+I) = 0$.

But, if $det(A+I) = 0 \Rightarrow (A+I)Y = 0 \Rightarrow AY = -Y$.

So $Y\neq X$ because if $A_{ij}>0 \quad and \quad x_i > 0 \Rightarrow AX \neq -X$

So, $det(A-I) = 0 \Rightarrow$ There is a vector Z that $(A-I)Z = 0 \Rightarrow AZ = Z$.

So, I need conclude that Z = X.

• How does $\det(A+I)=0$ imply $(A+I)X=0$? Jan 10, 2016 at 0:24
• so, I can't do this? I thought about eigenvectors and eigenvalues. If $det(A+I) = 0 \Rightarrow -1$ is a eigenvector. So, $AX = -X$. Isn't correct? Jan 10, 2016 at 0:32
• This would imply $-1$ is an eigenvalue, but I see no reason to conclude that $X$ is the corresponding eigenvector. In fact, $X$ is a matrix by definition. Jan 10, 2016 at 0:35
• Looks like $A$ is $I$. Jan 10, 2016 at 0:46
• Yeah, you have the reason M10687. Can you think one solution for this? I need conclude this for january 12 (three days for conclude :] ). Jan 10, 2016 at 0:53

More generally, let $A$ be a primitive matrix ($A$ is non-negative and some power of $A$ is a positive matrix) with spectral radius $r$ and $k\in\mathbb{N}$. If there is $x>0$ s.t. $A^kx=sx$, then $s=r^k$ and $Ax=rx$.

Proof. Note that $A^k$ is also primitive and then $\not=0$; thus, necessarily $s>0$ and we may assume that $s=1$. Using the Perron theorem, cf. https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem , since $x>0$, necessarily $1=\rho(A^k)=(\rho(A))^k$ and $r=1$. Note that $x\in \ker(A^k-I)=\ker(A-I)\bigoplus_{\omega;\omega^k=1,\omega\not=1} \ker(A-\omega I)$. Yet, $A$ is aperiodic ($1$ is the sole eigenvalue of $A$ with modulus $1$); thus $\ker(A^k-I)=\ker(A-I)$ and $Ax=x$.

Here is the outline of an elementary proof. Let $$\mathbf 1$$ be all-one vector, $$D$$ be the diagonal matrix whose main diagonal is $$x$$ (so that $$D\mathbf1=x$$) and $$B=D^{-1}AD$$. Then $$A^2x=x$$ is equivalent to $$B^2\mathbf 1=\mathbf 1$$ and we want to prove that $$\color{red}{B\mathbf 1=\mathbf 1}$$.

Let $$y=B\mathbf 1$$. Hence it is a positive vector. There are two possibilities:

1. $$y$$ and $$\mathbf1$$ are linearly independent. Let $$v$$ be the positive multiple of $$y$$ such that its largest element is equal to $$1$$. From $$B^2\mathbf1=\mathbf1$$, we get $$B^2y=y$$ and in turn $$B^2v=v$$. Therefore $$B^2(\mathbf1-v)=\mathbf1-v.\tag{\dagger}$$ Argue that $$\mathbf1-v$$ is a nonnegative vector with both zero entries and positive entries. Hence explain why ($$\dagger$$) is impossible.
2. $$y$$ is a scalar multiple of $$\mathbf1$$. Explain why this multiple must be positive; hence explain why it is equal to $$1$$. Now $$B\mathbf1=y=\mathbf1$$ and we are done.

Consider the characteristic polynomial $p$ of $A$. By the Cayley-Hamilton theorem, $p(A)=0$, and thus $p(A)X=0$. If $p(x)=\sum_{i=0}^n c_ix^i$, this gives $\sum_{i=0}^n c_iA^iX=0$. As $A^2X=X$, the latter sum simplifies to $d_1AX+d_0X=0$ for some constants $d_0$ and $d_1\not=0$. This in turn implies that $AX=\lambda X$ for $\lambda=-d_0/d_1$. Now $X=A^2X=\lambda^2 X$, so $\lambda=\pm 1$, and since all entries of $A$ and $X$ are positive, it can't be that $\lambda=-1$.

• Why is $d_1$ nonzero? E.g. when $A=\operatorname{diag}(1,-1)$ and $X=(1,0)^T$. Then $d_1=d_0=0$. While it may be possible to prove that $d_1\ne0$ by using the assumptions that $A$ and $X$ are positive, as it currently stands, your proof does not sound right. Jan 10, 2016 at 8:27
• @ fpguy , you can show that $d_1\not=0$ as follows. $2d_1=p(1)-p(-1)=-p(-1)$ (since $1$ is an eigenvalue of $A$). If $p(-1)=0$, then $-1$ is also an eigenvalue of $A$, that is impossible (according to Perron theorem).
– user91684
Jan 10, 2016 at 19:42
• I am amazed that a false proof was rewarded with a green chevron and $5$ upvotes.
– user91684
Jan 10, 2016 at 19:50