# Wolfram Alpha doesn't give right eigenvalues?

I used WolframAlpha to calculate the eigenvectors of the matrix

$$A= \begin{bmatrix} 1-a & 0 & (1-a)b & 0\\ a & 0 & ab & 0\\ 0 & 1-a & (1-a)(1-b) & (1-a)b\\ 0 & a & a(1-b) & 1-(1-a)b \end{bmatrix}$$

for $a=0.35, b=0.5$ with the command

a=0.35, b=0.5, eigenvectors(({{1 - a, 0, (1 - a) b, 0}, {a, 0, a b, 0}, {0, 1 - a, (1 - a) (1 - b), (1 - a) b }, {0, a, a (1 - b),1- (1- a)b}}))

It gives what I think is the correct result in the result section: $(0.513344, 0.276416, 0.552832, 0.595357)$ (the vector of which I'm interested) but in the eigenvalue section there's $\lambda_1 = -1.3033$ and not $1$ as I expected. Also, in the eigenvectors section, there isn't the vector given in the result (or its multiple), what's up with that?

Other software give the eigenvalue $1$ and the same eigenvector. Here's the matrix for input there if somebody wants to check

0.650 0.000 0.325 0.000

0.350 0.000 0.175 0.000

0.000 0.650 0.325 0.325

0.000 0.350 0.175 0.675

I have checked and double-checked and still get this weird output from Wolfram Alpha.

PS. If I change the command to eigenvalues, then it gives $1.$ and also other eigenvalues correctly (up to a precision). So clearly this is an error in WA.

• why should the result to be not right? – Dr. Sonnhard Graubner Jun 22 '17 at 17:05
• i can compute the Eigenvalues with Maple – Dr. Sonnhard Graubner Jun 22 '17 at 17:05
• When I enter merely a=0.35, b=0.5, {{1 - a, 0, (1 - a) b, 0}, {a, 0, a b, 0}, {0, 1 - a, (1 - a) (1 - b), (1 - a) b }, {0, a, a (1 - b),1- (1- a)b}} into WA, it lists an eigenvalue of $1$. So it can compute the correct eigenvalue, but somehow the "eigenvector" function call doesn't behave as expected. – Thomas Andrews Jun 22 '17 at 17:09
• Probably nobody here can answer the "what's up with that" question - just report it to Wolfram Alpha. – Thomas Andrews Jun 22 '17 at 17:12
• @Dr.SonnhardGraubner Then why do you think the correct values are helpful? The question is, why does WA get it wrong, and presumably, how to get WA to do the right thing. – Thomas Andrews Jun 22 '17 at 17:21

## 1 Answer

This is too long for a comment.

When we look at the characteristic polynomial given by Wolfram Alpha when asking for the eigenvectors (entering $a=\frac 7 {20}$ and $b=\frac 12$), it is reported to be $$\lambda^4+\frac{9}{2} \sqrt{\frac{13}{7}} \lambda^3+\frac{1355 }{588}\lambda^3-\frac{1501}{588} \sqrt{\frac{13}{7}} \lambda^2-\frac{845}{147}\lambda^2+\frac{319 }{196}\lambda-\frac{319 }{98}\sqrt{\frac{13}{7}}$$ the solutions of which being effectively $$\{ -9.43808, 1.15923, -0.0790194-0.631825 i, -0.0790194+0.631825 i\}$$ but they are not the same as those obtained when entering $a=0.35$ and $b=0.5$.

The characteristic polynomial should be $$\lambda ^4-\frac{33 }{20}\lambda ^3+\frac{559 }{800}\lambda ^2-\frac{39 }{800}\lambda$$ obtained from the general case and replacing $a$ and $b$ by their rational values; the solutions of this last polynomial being $$\left\{ 0, 1, \frac{13-\sqrt{91}}{40} , \frac{13+\sqrt{91}}{40} \right\}$$ More than strange, for sure.