Density matrix and von Neumann entropy - why does basis matter?

[Eigenentity]

Density matrix and von Neumann entropy -- why does basis matter?

I'm very confused by why I'm unable to correctly compute the von Neumann entropy

$$S = - \mathrm{Tr}(\rho \log_2{\rho})$$

for the pure state

$$| \psi \rangle = \left(|0\rangle + |1\rangle\right)/ \sqrt 2$$

Now, clearly the simplest thing to do is to express $$|\psi\rangle$$ in the $$|+\rangle,|-\rangle$$ basis, where it's clear that

$$\rho = |+\rangle\langle+|$$

In this basis, $$S = 0$$ as we'd expect for a pure state.

What I can't fathom (and I'm sure I'm missing something really obvious) is why if I evaluate the entropy in the $$|0\rangle,|1\rangle$$ basis,

$$\rho = \begin{pmatrix}<br /> 1/2 & 1/2 \\ <br /> 1/2 & 1/2 <br /> \end{pmatrix}$$

and thus

$$S = - \mathrm{Tr} \left( \begin{pmatrix}<br /> 1/2 & 1/2 \\ <br /> 1/2 & 1/2 <br /> \end{pmatrix} \begin{pmatrix}<br /> -1 & -1 \\ <br /> -1 & -1 <br /> \end{pmatrix} \right) = - \mathrm{Tr} \begin{pmatrix}<br /> -1 & -1 \\ <br /> -1 & -1 <br /> \end{pmatrix} = 2$$

Do I have to use the diagonal basis!? If not, what idiotic mistake am I making?

 

[Eigenentity]

To anyone wondering. No, basis does not matter. The error in the above reasoning is the assumption that the logarithm of a matrix is performed elementwise. This is true for diagonal matrices, but not general ones.

More concretely:

$$\log_2 \begin{pmatrix}<br /> 1/2 & 1/2 \\ <br /> 1/2 & 1/2 <br /> \end{pmatrix} \neq \begin{pmatrix}<br /> -1 & -1 \\ <br /> -1 & -1<br /> \end{pmatrix}$$

Gah! Well, glad that's sorted.