1 Attachment(s)

Question about Property of Spectral Radius (Modified)

My Claim below, if proved, would be extremely useful (in economics).

Simulations seem to confirm my claim, but I can't prove it. Please help me!

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.

Re: Question about Property of Spectral Radius (Modified)

Quote:

Originally Posted by

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

Simulations seem to confirm my claim, but I can't prove it. Please help me!

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

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

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

Thus $\displaystyle \rho(S_X)>1$ for all X satisfying the given constraints, but $\displaystyle \rho(A)>1$ also. So (2) holds, but not (1).

Re: Question about Property of Spectral Radius (Modified)

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.

Re: Question about Property of Spectral Radius (Modified)

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 $\displaystyle \varepsilon$ to the off-diagonal elements of A, to get the matrix $\displaystyle A_\varepsilon = \begin{bmatrix}2&\varepsilon\\ \varepsilon&\frac12\end{bmatrix},$ with all the entries strictly positive. The possible values for the corresponding matrix $\displaystyle X_\varepsilon$ are $\displaystyle \begin{bmatrix}\frac12(1-\varepsilon q)&p\\q&2(1-\varepsilon p)\end{bmatrix}.$ Then $\displaystyle S_{A_\varepsilon} = \begin{bmatrix}4&\varepsilon^2\\ \varepsilon^2&\frac14\end{bmatrix}$ and $\displaystyle S_{X_\varepsilon} = \begin{bmatrix}\frac14(1-\varepsilon q)^2&p^2\\q^2&4(1-\varepsilon p)^2\end{bmatrix}.$

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

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

Re: Question about Property of Spectral Radius (Modified)

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 $\displaystyle \varepsilon$ to the off-diagonal elements of A, to get the matrix $\displaystyle A_\varepsilon = \begin{bmatrix}2&\varepsilon\\ \varepsilon&\frac12\end{bmatrix},$ with all the entries strictly positive. The possible values for the corresponding matrix $\displaystyle X_\varepsilon$ are $\displaystyle \begin{bmatrix}\frac12(1-\varepsilon q)&p\\q&2(1-\varepsilon p)\end{bmatrix}.$ Then $\displaystyle S_{A_\varepsilon} = \begin{bmatrix}4&\varepsilon^2\\ \varepsilon^2&\frac14\end{bmatrix}$ and $\displaystyle S_{X_\varepsilon} = \begin{bmatrix}\frac14(1-\varepsilon q)^2&p^2\\q^2&4(1-\varepsilon p)^2\end{bmatrix}.$

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

I don't know whether this is significant for the applications to economics, but mathematically it seems much more plausible that the converse implication $\displaystyle (1)\Rightarrow(2)$ 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, $\displaystyle (1)\Rightarrow(2)$ 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 $\displaystyle (1)\Rightarrow(2)$ is true? If so (if not) how to prove(disprove) it?

I cannot find an appropriate word about how I thank you.

Re: Question about Property of Spectral Radius (Modified)

Dear Opalg,

I thought about your example a little more and I found that adding a small positive number $\displaystyle \varepsilon$ makes a huge difference. In your example, set $\displaystyle p=1/\varepsilon$, and $\displaystyle q=\varepsilon^n$. Then you can verify that the characteristic function of $\displaystyle S_{X_\varepsilon}$ is given by: $\displaystyle \xi^2-\frac14(1-\varepsilon^{n+1})^2 \xi -\varepsilon^{2n-2}$. Hence, as $\displaystyle n$ gets larger, the maximum eigenvalue of $\displaystyle S_{X_\varepsilon}$ converges to $\displaystyle \frac14$.

Therefore, for some $\displaystyle S_{X_\varepsilon}$, $\displaystyle \rho(S_{X_\varepsilon})<1$ while $\displaystyle \rho(S_{A_\varepsilon})>1$. This is in contrast to the case with $\displaystyle \varepsilon=0$. Hence, $\displaystyle (2)\Rightarrow(1)$ may still be true.

What you do think about this?

Re: Question about Property of Spectral Radius (Modified)

Quote:

Originally Posted by

**sc719** Dear Opalg,

I thought about your example a little more and I found that adding a small positive number $\displaystyle \varepsilon$ makes a huge difference. In your example, set $\displaystyle p=1/\varepsilon$, and $\displaystyle q=\varepsilon^n$. Then you can verify that the characteristic function of $\displaystyle S_{X_\varepsilon}$ is given by: $\displaystyle \xi^2-\frac14(1-\varepsilon^{n+1})^2 \xi -\varepsilon^{2n-2}$. Hence, as $\displaystyle n$ gets larger, the maximum eigenvalue of $\displaystyle S_{X_\varepsilon}$ converges to $\displaystyle \frac14$.

Therefore, for some $\displaystyle S_{X_\varepsilon}$, $\displaystyle \rho(S_{X_\varepsilon})<1$ while $\displaystyle \rho(S_{A_\varepsilon})>1$. This is in contrast to the case with $\displaystyle \varepsilon=0$. Hence, $\displaystyle (2)\Rightarrow(1)$ may still be true.

What you do think about this?

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 $\displaystyle p$ 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.

Re: Question about Property of Spectral Radius (Modified)

$\displaystyle a11,..,a22 $ are simply non-zero (positive or negative) real numbers, thus the elements of $\displaystyle A = \begin{bmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix}$ can be positive or negative. But the elements of $\displaystyle S_A = \begin{bmatrix}a_{11}^2&a_{12}^2\\ a_{21}^2&a_{22}^2\end{bmatrix}$ are strictly positive. Likewise, $\displaystyle X = \begin{bmatrix}x_{11}&x_{21}\\ x_{12}&x_{22}\end{bmatrix}$ can have positive or negative or even zero, while the elements of $\displaystyle S_X = \begin{bmatrix}x_{11}^2&x_{21}^2\\ x_{12}^2&x_{22}^2\end{bmatrix}$ are non-negative.

My example above indeed sets $\displaystyle x_{22}=2(1-\varepsilon p)=0$. And this is admissible. But by setting $\displaystyle p=(1-\varepsilon^{n+1})/\varepsilon$, and $\displaystyle q=\varepsilon^{n}$, the elements of $\displaystyle X$ are all non-zero and the characteristic fucntion of $\displaystyle S_X$ is given by

$\displaystyle \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, $\displaystyle \rho(S_X) $ converges to 1/4 quickly.

This is happening as $\displaystyle \rho(S_A) $ is closed to 4. So, I guess that as long as $\displaystyle \rho(S_A)>1 $, I can make a example such that for some $\displaystyle X$ (subject to the constraints in the original question), $\displaystyle \rho(S_X)<1 $.

I am glad to see that you agree with me in that my claim is possibly true. So back to my question again. My claim seems true, but I don't know how to prove it.

The parameter space of $\displaystyle \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 $\displaystyle \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.

Please help me!