Why $0^0=I$?
I'd tried prove that considering $N^0$ where N is a Nilpotent matrix and then using the Cayley -Hamilton theorem
Thanks in advance.
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You could have a look at a similar question for numbers. Some of the arguments given there make sense for matrices, too. – Martin Sleziak Sep 27 '15 at 06:19
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The same caveats as for the exponential of real numbers apply here. If exponentiation means repeated multiplication, then $A^0 = I$ is the base case for all $A$. Exponentiation by a continuous real parameter, on the other hand, should insist that $\mathbf{0}^0$ is undefined.
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Just like the empty sum is defined to be the additive identity $0$, the empty product is usually defined to be the multiplicative identity $1$. For matrices, the multiplicative identity is the identity matrix.
RghtHndSd
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0^0 must be undefined because here in case of matrices if we take matrix's power than it implies the product of that matrix that power times and it does not make sense for the zero element's power zero I.e. Zeros' power zero , which is undefined.
