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Bookmark or Share- If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$?Short Answer . Yes AA-1 = A-1 A = I when the Det(A) $\neq$ 0 and A is a square matrix.. Long Answer . A matrix is basically a linear transformation applied to some space.
- General term formula of series 1/1 + 1/2 + 1/3 ... +1/nI can give you a good approximation if you would prefer: $$\ln(n+1)\le\sum_{i=1}^n\frac1i\le\ln(n)+1$$ This is a rather tight upper limit and lower limit you can use to approximate your answer.
- abstract algebra - Prove that 1+1=2 - Mathematics Stack ExchangeThe main reason that it takes so long to get to $1+1=2$ is that Principia Mathematica starts from almost nothing, and works its way up in very tiny, incremental steps. The work of G. Peano shows that it's not hard to produce a useful set of axioms that can prove 1+1=2 much more easily than Whitehead and Russell do.
- What does $QAQ^{-1}$ actually mean? - Mathematics Stack Exchange105 1 1 silver badge 5 5 bronze badges $\endgroup$ 2. 1 $\begingroup$ Not an answer, just a comment. This ...
- What is the value of $1^i$? - Mathematics Stack ExchangeIf we take a branch of the logarithm where $\log(1) = 2\pi i$ then $1^{1/2} = \exp(\pi i) = -1$ which is not equal to $1^1 = \exp(2\pi i) = 1$. Complex exponentiation is not the same function as real exponentiation; they only agree for appropriate branches of the logarithm, and then that agreement has to be proved, it is not a definition.
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