• Jul 18th 2011, 09:07 PM
sc719
My Claim below, if proved, would be extremely useful (in economics).

For given (real valued) scalars, (a11,..a22), (real-valued) x11,..,x22 must satisfy the following restrictions.
a11*x11+a12*x12=1,
a21*x21+a22*x22=1,

Define the following matrices.
S_A=[a11^2 a12^2;a21^2 a22^2]
S_X=[x11^2 x21^2;x12^2 x22^2] : ((2,1)-th element is x12^2, not x21^2)

Let rho(B)=maximum eigenvalue of n by n matrix, B in absolute value.

My Claim : The following assertions are equivalent:
(1) rho(S_A)<1
(2) rho(S_X)>1 for any S_X with x11,..x22 obeying the constraints above.

My Claim in general setup is given in the half-page long attached pdf file.
• Jul 19th 2011, 01:53 PM
Opalg
Quote:

Originally Posted by sc719
My Claim below, if proved, would be extremely useful (in economics).

For given (real valued) scalars, (a11,..a22), (real-valued) x11,..,x22 must satisfy the following restrictions.
a11*x11+a12*x12=1,
a21*x21+a22*x22=1,

Define the following matrices.
S_A=[a11^2 a12^2;a21^2 a22^2]
S_X=[x11^2 x21^2;x12^2 x22^2] : ((2,1)-th element is x12^2, not x21^2)

Let rho(B)=maximum eigenvalue of n by n matrix, B in absolute value.

My Claim : The following assertions are equivalent:
(1) rho(S_A)<1
(2) rho(S_X)>1 for any S_X with x11,..x22 obeying the constraints above.

My Claim in general setup is given in the half-page long attached pdf file.

Unless I am misunderstanding something, I don't see how this can be true.

Write $A = \begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}$ and $X = \begin{bmatrix}x_{11}&x_{21}\\ x_{12}&x_{22}\end{bmatrix}.$ Suppose that we take $A = \begin{bmatrix}2&0\\ 0&\frac12\end{bmatrix}.$ The possible values of X satisfying the given restrictions are $X = \begin{bmatrix}\frac12&p\\ q&2\end{bmatrix},$ where p and q are arbitrary. Then $S_A = \begin{bmatrix}4&0\\ 0&\frac14\end{bmatrix}$ and $S_X = \begin{bmatrix}\frac14&p^2\\ q^2&4\end{bmatrix}.$

The eigenvalues of a 2x2 matrix $T = \begin{bmatrix}a&b\\ c&d\end{bmatrix}$ are given by $\lambda = \tfrac12\bigl(\tau \pm \sqrt{\tau^2-4\delta}\bigr) = \tfrac12\bigl(a+d+\sqrt{(a-d)^2+4bc}\bigr),$ where $\tau = a+d$ is the trace of T, and $\delta = ad-bc$ is its determinant. If the eigenvalues are real, then the spectral radius (the larger of their absolute values) is given by $\rho(T) = \tfrac12\bigl(|a+d|+\sqrt{(a-d)^2+4bc}\bigr).$

Applying that to the above matrices $S_A$ and $S_X$, you see that $\rho(A) = 4$ and (noting that $S_X$ has real eigenvalues for all values of p and q) $\rho(S_X) = \tfrac12\Bigl(\tfrac{17}4 + \sqrt{\bigl(\tfrac{15}4\bigr)^2 + 4p^2q^2}\Bigr) \geqslant \tfrac12\bigl(\tfrac{17}4 + \tfrac{15}4\bigr) = 4.$

Thus $\rho(S_X)>1$ for all X satisfying the given constraints, but $\rho(A)>1$ also. So (2) holds, but not (1).
• Jul 24th 2011, 07:52 AM
sc719
Dear Opalg,

Thank you so much for your counterexample to my claim! You are absolutely right and more importantly, your comment sharpens my question.

I assumed implicitly that a11,..,a22 are all non-zero(A11,.., A22 are all non-singular in the attachment.).
And I would like to ask you once more whether my claim is true, and if so (if not), how to prove(disprove) it, when a11,...,a22 are all non-zero.

Thank you very much in advance.
• Jul 25th 2011, 08:58 AM
Opalg
Quote:

Originally Posted by sc719
I assumed implicitly that a11,..,a22 are all non-zero(A11,.., A22 are all non-singular in the attachment.).
And I would like to ask you once more whether my claim is true, and if so (if not), how to prove(disprove) it, when a11,...,a22 are all non-zero.

It is still not true that (2) implies (1), even if all the elements of A are strictly positive. In fact, suppose that you add a small positive number $\varepsilon$ to the off-diagonal elements of A, to get the matrix $A_\varepsilon = \begin{bmatrix}2&\varepsilon\\ \varepsilon&\frac12\end{bmatrix},$ with all the entries strictly positive. The possible values for the corresponding matrix $X_\varepsilon$ are $\begin{bmatrix}\frac12(1-\varepsilon q)&p\\q&2(1-\varepsilon p)\end{bmatrix}.$ Then $S_{A_\varepsilon} = \begin{bmatrix}4&\varepsilon^2\\ \varepsilon^2&\frac14\end{bmatrix}$ and $S_{X_\varepsilon} = \begin{bmatrix}\frac14(1-\varepsilon q)^2&p^2\\q^2&4(1-\varepsilon p)^2\end{bmatrix}.$

If $\varepsilon$ is small enough, then $\rho(S_{A_\varepsilon})$ and $\rho(S_{X_\varepsilon})$ will be close to $\rho(S_A)$ and $\rho(S_X)$ respectively, so you will still get a counterexample to $(2)\Rightarrow(1).$

I don't know whether this is significant for the applications to economics, but mathematically it seems much more plausible that the converse implication $(1)\Rightarrow(2)$ might be true. At any rate, I think it would not be nearly as easy to concoct a counterexample.
• Jul 25th 2011, 10:25 AM
sc719
Quote:

Originally Posted by Opalg
It is still not true that (2) implies (1), even if all the elements of A are strictly positive. In fact, suppose that you add a small positive number $\varepsilon$ to the off-diagonal elements of A, to get the matrix $A_\varepsilon = \begin{bmatrix}2&\varepsilon\\ \varepsilon&\frac12\end{bmatrix},$ with all the entries strictly positive. The possible values for the corresponding matrix $X_\varepsilon$ are $\begin{bmatrix}\frac12(1-\varepsilon q)&p\\q&2(1-\varepsilon p)\end{bmatrix}.$ Then $S_{A_\varepsilon} = \begin{bmatrix}4&\varepsilon^2\\ \varepsilon^2&\frac14\end{bmatrix}$ and $S_{X_\varepsilon} = \begin{bmatrix}\frac14(1-\varepsilon q)^2&p^2\\q^2&4(1-\varepsilon p)^2\end{bmatrix}.$

If $\varepsilon$ is small enough, then $\rho(S_{A_\varepsilon})$ and $\rho(S_{X_\varepsilon})$ will be close to $\rho(S_A)$ and $\rho(S_X)$ respectively, so you will still get a counterexample to $(2)\Rightarrow(1).$

I don't know whether this is significant for the applications to economics, but mathematically it seems much more plausible that the converse implication $(1)\Rightarrow(2)$ might be true. At any rate, I think it would not be nearly as easy to concoct a counterexample.

Dear Opalg,

You saved my research time so much and I greatly appreciate your comments.
My claim may be no more be true. But as you conjecture, $(1)\Rightarrow(2)$ might be still true. This sufficiency of (1) for (2) is much more important in economics than its necessity.

So, may I ask you once more a big favor? Do you think that $(1)\Rightarrow(2)$ is true? If so (if not) how to prove(disprove) it?

I cannot find an appropriate word about how I thank you.
• Jul 26th 2011, 03:47 AM
sc719
Dear Opalg,

I thought about your example a little more and I found that adding a small positive number $\varepsilon$ makes a huge difference. In your example, set $p=1/\varepsilon$, and $q=\varepsilon^n$. Then you can verify that the characteristic function of $S_{X_\varepsilon}$ is given by: $\xi^2-\frac14(1-\varepsilon^{n+1})^2 \xi -\varepsilon^{2n-2}$. Hence, as $n$ gets larger, the maximum eigenvalue of $S_{X_\varepsilon}$ converges to $\frac14$.
Therefore, for some $S_{X_\varepsilon}$, $\rho(S_{X_\varepsilon})<1$ while $\rho(S_{A_\varepsilon})>1$. This is in contrast to the case with $\varepsilon=0$. Hence, $(2)\Rightarrow(1)$ may still be true.

• Jul 26th 2011, 01:19 PM
Opalg
Quote:

Originally Posted by sc719
Dear Opalg,

I thought about your example a little more and I found that adding a small positive number $\varepsilon$ makes a huge difference. In your example, set $p=1/\varepsilon$, and $q=\varepsilon^n$. Then you can verify that the characteristic function of $S_{X_\varepsilon}$ is given by: $\xi^2-\frac14(1-\varepsilon^{n+1})^2 \xi -\varepsilon^{2n-2}$. Hence, as $n$ gets larger, the maximum eigenvalue of $S_{X_\varepsilon}$ converges to $\frac14$.
Therefore, for some $S_{X_\varepsilon}$, $\rho(S_{X_\varepsilon})<1$ while $\rho(S_{A_\varepsilon})>1$. This is in contrast to the case with $\varepsilon=0$. Hence, $(2)\Rightarrow(1)$ may still be true.

Yes, you're quite right. Sorry to have misled you. Just to be clear about the statement of the problem, the elements of A must be strictly positive, but the elements of X need not be. In your example, you are taking $p$ in a way that makes one of the elements of X zero, but I guess that X could even have negative elements. Is that right?

In all the examples that I can construct, it does seem that conditions (1) and (2) are equivalent. But I don't see any way to prove the implication in either direction.
• Jul 26th 2011, 10:13 PM
sc719
$a11,..,a22$ are simply non-zero (positive or negative) real numbers, thus the elements of $A = \begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}$ can be positive or negative. But the elements of $S_A = \begin{bmatrix}a_{11}^2&a_{12}^2\\ a_{21}^2&a_{22}^2\end{bmatrix}$ are strictly positive. Likewise, $X = \begin{bmatrix}x_{11}&x_{21}\\ x_{12}&x_{22}\end{bmatrix}$ can have positive or negative or even zero, while the elements of $S_X = \begin{bmatrix}x_{11}^2&x_{21}^2\\ x_{12}^2&x_{22}^2\end{bmatrix}$ are non-negative.
My example above indeed sets $x_{22}=2(1-\varepsilon p)=0$. And this is admissible. But by setting $p=(1-\varepsilon^{n+1})/\varepsilon$, and $q=\varepsilon^{n}$, the elements of $X$ are all non-zero and the characteristic fucntion of $S_X$ is given by
$\xi^2-(\frac{1}{4}(1-\varepsilon^{n+1})^2+4 \varepsilon^{2n+2})\xi-(1-\varepsilon^{n+1})^2 (1-\varepsilon^4) \varepsilon^{2n-2} =0$. Thus, even n=2 or larger, $\rho(S_X)$ converges to 1/4 quickly.
This is happening as $\rho(S_A)$ is closed to 4. So, I guess that as long as $\rho(S_A)>1$, I can make a example such that for some $X$ (subject to the constraints in the original question), $\rho(S_X)<1$.
The parameter space of $\rho(S_A)<1$ seems a convex hull (ball?) in the Euclidean space. Knowing One subset in the Euclidean space is equivalent to knowing the complement of the subset. But Knowing the set $\rho(S_A)<1$ is much easier than knowing its complement. Maybe I should resort to compeletly a different mathematical tool to prove my claim.