Try this
y0 = (u - 1)^(1/2);
jacob = {{-1, 0}, {y0, -2 y0^2}} // Eigensystem

MatrixForm
got in the way of your computation.
To be able to use MatrixForm
and not have it stop your computation, one option is
y0 = (u - 1)^(1/2);
(jacob = {{-1, 0}, {y0, -2 y0^2}}) // MatrixForm

Eigensystem[jacob]

Added verification by hand per comment. Mathematica result is correct.
$$
A=
\begin{pmatrix}
-1 & 0\\
\sqrt{u-1} & -2\left( u-1\right)
\end{pmatrix}
$$
Hence
$$
\left\vert A-\lambda I\right\vert =
\begin{vmatrix}
-1-\lambda & 0\\
\sqrt{u-1} & -2\left( u-1\right) -\lambda
\end{vmatrix}
$$
Therefore $\left\vert A-\lambda I\right\vert =0$ gives
\begin{align*}
\left( -1-\lambda\right) \left( -2\left( u-1\right) -\lambda\right) &
=0\\
2u-\lambda+2u\lambda+\lambda^{2}-2 & =0\\
\lambda^{2}+\lambda\left( 2u-1\right) -2+2u & =0
\end{align*}
Using the quadratic formula $\lambda=\frac{-b}{2a}\pm\frac{1}{2a}\sqrt
{b^{2}-4ac}$ the roots are
\begin{align*}
\lambda & =\frac{-\left( 2u-1\right) }{2}\pm\frac{1}{2}\sqrt{\left(
2u-1\right) ^{2}-4\left( -2+2u\right) }\\
& =\frac{-2u+1}{2}\pm\frac{1}{2}\sqrt{\left( 2u-3\right) ^{2}}\\
& =\frac{-2u+1}{2}\pm\frac{1}{2}\left( 2u-3\right)
\end{align*}
Hence
\begin{align*}
\lambda_{1} & =\frac{-2u+1}{2}+\frac{1}{2}\left( 2u-3\right) \\
& =\frac{-2u}{2}+\frac{1}{2}+u-\frac{3}{2}\\
& =-u+\frac{1}{2}+u-\frac{3}{2}\\
& =-1
\end{align*}
And
\begin{align*}
\lambda_{2} & =\frac{-2u+1}{2}-\frac{1}{2}\left( 2u-3\right) \\
& =\frac{-2u}{2}+\frac{1}{2}-u+\frac{3}{2}\\
& =-u+\frac{1}{2}-u+\frac{3}{2}\\
& =-2u+2\\
& =-2\left( u-1\right)
\end{align*}