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Let $$A = \begin{pmatrix} 0.6 & 0.1 & 0.1\\ 0.1 & 0.8 & 0.2\\ 0.3 & 0.1 & 0.7 \end{pmatrix}$$

I want to find the eigenvalues of this matrix. Because this is a markov matrix, I know that $\lambda_1 = 1$. Furthermore, I know that the remaining eigenvalues $\lambda_2$ and $\lambda_3$ should be less than one (their magnitudes should be less than one, at least) because all the entries in $A$ are positive.

I found the characteristic polynomial, but it is an ugly polynomial. Is there a better/easier way to determine the remaining two eigenvalues? Maybe by using the information I already have that $\lambda_1 = 1$?

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    $\begingroup$ A $3\times3$ matrix creates a cubic characteristic equation. Thus if you know one of the eigenvalues ($\lambda_1 = 1$) then you can use polynomial long division (or synthetic division) to find a quadratic polynomial and solve for the final two eigenvalues. $\endgroup$ – Jared Sep 30 '14 at 6:24
  • $\begingroup$ I did contemplate dividing my characteristic polynomial by $(\lambda - 1)$. $\endgroup$ – s7even Sep 30 '14 at 6:26
  • $\begingroup$ Perhaps you should show the results of that...note that you don't need the decimals...you can just as easily find the eigenvalues for $\begin{pmatrix}6&1&1\\1&8&1\\3&1&7\end{pmatrix}$ (I believe--I'm sure someone will correct me if I'm wrong about that). $\endgroup$ – Jared Sep 30 '14 at 6:28
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    $\begingroup$ @Jared well, the eigenvalues of $10A$ are ten times larger than the eigenvalues of $A$. $\endgroup$ – 5xum Sep 30 '14 at 6:33
  • $\begingroup$ @5xum OK, yeah you're right, because you would have: $Ax - \lambda x = 0$ vs. $\alpha Ax - \alpha\lambda x = 0$. $\endgroup$ – Jared Sep 30 '14 at 6:34
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  1. Calculate $p(\lambda) = \det(A-\lambda I)$, the characteristic polynomial
  2. You know that $p(1)=0$, so you transform $p(\lambda)$ to $p(\lambda)=(\lambda-1)\cdot r(\lambda)$ where $r$ is a second order polynomial (you can find it via Horner's algorithm)
  3. Find the roots of $r$.
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  • $\begingroup$ If the rows also added up to 1 wouldn't that be a doubly-stochastic matrix, not a markov matrix? $\endgroup$ – s7even Sep 30 '14 at 6:44
  • $\begingroup$ @s7even Well, it is stochastic if either the rows or the columns add to $1$. In your case, NEITHER is true. $\endgroup$ – 5xum Sep 30 '14 at 6:46
  • $\begingroup$ I see. On the 2nd row, 3rd column, I typed 0.1 instead of 0.2. My mistake. $\endgroup$ – s7even Sep 30 '14 at 6:47
  • $\begingroup$ @s7even Yes, it's stochastic now. And my answer is a sure way to get the solution. $\endgroup$ – 5xum Sep 30 '14 at 6:50
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So I'm going to go with integers here (for no real reason other than it's easier to type--so my eigenvalues will be $10$ times the actual eigenvalues--per @5xum's comment).

$$ (6 - \lambda)((8 - \lambda)(7 - \lambda) - 2) - ((7 - \lambda) - 6) + (1 - 3(8 - \lambda) = 0 \\ (6 - \lambda)(56 - 15\lambda + \lambda^2 - 2) - 7 + \lambda + 6 + 1 - 24 + 3\lambda = 0 \\ 324 - 90\lambda + 6\lambda^2 - 54\lambda + 15\lambda^2 - \lambda^3 - 24 + 4\lambda = 0 \\ -\lambda^3 + 21\lambda^2-140\lambda + 300 = 0 $$

Indeed this has a zero at $\lambda = 10\cdot 1$. Now we attempt to find:

$$ \lambda^3 - 21\lambda^2 + 140\lambda - 300 = (\lambda - 10)(a\lambda^2 + b\lambda + c) \\ \lambda^3 - 21\lambda^2 + 140\lambda - 300 = a\lambda^3 + (b - 10a)\lambda^2 + (c - 10b)\lambda - 10c $$

Yielding:

\begin{align} a =&\ 1 &&\longrightarrow &&a&&=& 1 \\ b - 10a =&\ -21 &&\longrightarrow &&b = -21 + 10 &&=&\ -11 \\ c - 10b =&\ 140 &&\longrightarrow &&c = 140 - 10\cdot11 &&=&\ 30 \\ -10c =&\ -300 &&\longrightarrow&& c = \frac{300}{10} &&=&\ 30 \end{align}

We now have:

$$ \lambda^3 - 21\lambda^2 + 140\lambda - 300 = (\lambda - 10)(\lambda^2 - 11\lambda + 30) $$

Using the quadratic equation, this yields:

$$ \lambda = \frac{11 \pm \sqrt{121 - 120}}{2} = \frac{11 \pm 1}{2} = \{6, 5\} $$

Since these represent 10 times the original eigenvalue, the other two (actual) eigenvalues are $\lambda = \{1, 0.6, 0.5\}$

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