Why is $\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right)^0\ne1$? Why is
$\left( \begin{array}{cc}  1 & 1 \\  1 & 1 \\ \end{array} \right)^0=\left( \begin{array}{cc}  \frac{1}{2} & \frac{1}{2} \\  \frac{1}{2} & \frac{1}{2} \\ \end{array} \right)$
This result me returns Mathematica on this command
Limit[MatrixFunction[#^s &, ( { {1, 1}, {1, 1} } )], s -> 0, Direction -> "FromAbove"]

Result:
{{1/2, 1/2}, {1/2, 1/2}}

Should not anything rised to zeroth power be $1$?
 A: As a limit it is like $0^0$. So you shouldn't write $A^0$. You can diagonalize it, it has eigenvalues 2,0 with corresponding eigenvectors,
$$E = \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}$$
So now we can look at the nth power as,
$$A^n = E^{-1} \begin{bmatrix} 2^n & 0 \\ 0 & 0^n \end{bmatrix} E$$
Taking the limit of this as $n \to 0^+$ now we have $2^0=1$ in the upper entry and we end up with your result.
A: I would argue that
$$\begin{bmatrix}1 & 1 \\ 1 &  1 \end{bmatrix}^0$$
could be any symmetric $2×2$ matrix with an eigenvector $\begin{bmatrix}1 \\ 1 \end{bmatrix}$ having eigenvector $1$, to wit:
$$\begin{bmatrix}a & 1-a \\ 1-a &  a \end{bmatrix}$$
Any such matrix multiplied by right or left by $\begin{bmatrix}1 & 1 \\ 1 &  1 \end{bmatrix}$ gives back $\begin{bmatrix}1 & 1 \\ 1 &  1 \end{bmatrix}$ and thus qualifies potentially as a zero power by the multiplicative definition of exponentiation.
This unsatisfactory result stems from the fact that all rows and columns of the matrix $\begin{bmatrix}1 & 1 \\ 1 &  1 \end{bmatrix}$ are multiples of a specific vector, or more generally that they are linearly dependent, or getting to the crux of the matter, the input matrix is singular!  Just as with numbers running the multiplicative recursion backwards fails to give $0^0=1$ because $0$ lacks a multiplicative inverse, so it is with matrices.
In the case of $0^0$ there are definitions of exponentiation (with limited domains on the exponent) that effectively force $0^0=1$, but matrices with their greater degree of freedom are harder to pin down and give less predictable results.  Something in the way Mathematica defines matrix exponentiation, or perhaps the limit process used in the program,, led it to a specific value of an intrinsically undefined zero-power of a singular matrix.
A: $A^0$ is the identity matrix. This then satisfies the rule
$$A^{m+n}=A^mA^n$$
which would not be true with your value for $A^0$.
Your question should be why is the limit not $I$?
