Wolfram Alpha doesn't give right eigenvalues?

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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.

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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.