1
$\begingroup$

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.

$\endgroup$
8
  • 1
    $\begingroup$ How does $\det(A+I)=0$ imply $(A+I)X=0$? $\endgroup$
    – M10687
    Jan 10, 2016 at 0:24
  • $\begingroup$ 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? $\endgroup$ Jan 10, 2016 at 0:32
  • 1
    $\begingroup$ 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. $\endgroup$
    – M10687
    Jan 10, 2016 at 0:35
  • $\begingroup$ Looks like $A$ is $I$. $\endgroup$
    – NoChance
    Jan 10, 2016 at 0:46
  • $\begingroup$ Yeah, you have the reason M10687. Can you think one solution for this? I need conclude this for january 12 (three days for conclude :] ). $\endgroup$ Jan 10, 2016 at 0:53

3 Answers 3

4
$\begingroup$

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 $Ax=x$.

$\endgroup$
3
$\begingroup$

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.
$\endgroup$
1
$\begingroup$

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 $\lambda=-1$.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – user1551
    Jan 10, 2016 at 8:27
  • $\begingroup$ @ 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). $\endgroup$
    – user91684
    Jan 10, 2016 at 19:42
  • 1
    $\begingroup$ I am amazed that a false proof was rewarded with a green chevron and $5$ upvotes. $\endgroup$
    – user91684
    Jan 10, 2016 at 19:50

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.