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If one matrix whose determinant is equal to 0 which means it doesn't have an inverse. Then how is possible to find the value of the matrix to the power of 0 equal to identity matrix when multiplying the original matrix with something undefined?

Is it a math fluke, or I am missing some important information?

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This is a good question.

The reason we define $A^0 = I$ is so that the identity $$ A^{m+n} = A^mA^n $$ holds whenever $m$ and $n$ are nonnegative integers. Then we can evaluate $p(A)$ for any polynomial $p$, and sometimes even compute with power series to get things like $e^A$.

This all makes sense and is often useful whether or not $A$ is invertible. When it is, we extend the definition so that $A^{-n} = (A^{-1})^n$.

Ethan Bolker
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    What if $A$ is the zero matrix? Are you still defining $A^0$? For the one dimensional case, this would contradict the usual convention that $0^0$ is undefined. – Qudit Mar 29 '17 at 19:37
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    @Qudit $0^0$ is a special case. Probably best to leave it undefined. But for evaluating $p(A)$ for polynomial $p$ defining it to be $I$ makes everything work. – Ethan Bolker Mar 29 '17 at 19:40
  • It's more than just the zero matrix that this is an issue for. If $A = \mathrm{diag}(1, 0)$, then your definition still implies that $0^0 = 1$. In fact, the same is true of any matrix that is not invertible if we change the basis. – Qudit Mar 29 '17 at 19:44
  • @Qudit I don't think we have to debate this further. Any definition (or none) that allows the functional calculus to work is what's required. $e^{\text{diag}(1,0)}$ should come out to $\text{diag}(e,1)$. – Ethan Bolker Mar 29 '17 at 20:48
  • @EthanBolker I was reading Lawrence Perko (Page 13 of Differential Equations and Dynamical System) where he tries to show that the inverse of $e^T$ is $e^{-T}$, if I understand it correctly the corollary 2 uses the fact $e^Te^{-T} = e^0$ is the identity matrix since the summation has $T^0$ in the first term. So I think matrix 0 to the power 0 is also I. – Aditya P Feb 19 '21 at 08:23