$e^{0\times I}$ in Wolfram Alpha broken?

Question

How comes that

$$ e^{0_2} = e^{\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}} = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} = 1_2 $$

according to Wolfram Alpha? Shouldn't $e$ to the power of a square matrix consisting of all zeros be the identity matrix of the same size? By my reasoning, we have

$$ e^{0_2}\times e^{0_2} = e^{0_2 + 0_2} = e^{0_2} $$

(I know that $e^A\times e^B$ is not equal to $e^{A+B}$ in general, since $e^A$ and $e^B$ might not commute, but that is how I think of it in this case), but

$$ 1_2\times 1_2 = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix} = 2_2 \neq 1_2, $$

so we must have

$$ e^{0_2} \neq 1_2. $$

Is there something wrong with this reasoning?

Answer 1

If you double-check Wolfram, you'll see that you performed an elementwise operation:

If you instead use MatrixExp instead of e^, you will get the desired result (link):

Answer 2

An annoying thing that happens with matrix operations in software is the need to specify whether you mean "apply this operation matrix-wise" or "apply this operation element-wise". In particular, for $A=0_2$, $e^A$ (matrix-wise) is indeed $I_2$. Meanwhile the $B$ given by $B_{ij}=e^{a_{ij}}$ (element-wise) is $1_2$. As WA is warning you, it is interpreting your input as "matrix element-wise exponentiation", so you are going to end up with $B$ rather than $e^A$.

Answer 3

The problem is you told wolfram alpha to do

$$e^\hat{0}=\begin{pmatrix}e^0 & e^0 \\ e^0 & e^0\end{pmatrix}=\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$$

What you see seemed to think it would do though is

$$e^\hat{0}=I+\sum_{k=1}^\infty \frac{\hat{0}^k}{k!}=I=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$

I know in mathematica (not sure about wolfram though) that this is given by MatrixExp.