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- What does $QAQ^{-1}$ actually mean? - Mathematics Stack Exchange
1 $\begingroup$ When one thinks of matrix products like that, it's helpful to remember that matrices, unlike vectors, have two sets of bases: one for the domain and one for the range Thinking of applying a vector on to the right, we get that the transformation "unrotates" the vector, applies the original transformation in the original basis
- What is the value of $1^i$? - Mathematics Stack Exchange
There are infinitely many possible values for $1^i$, corresponding to different branches of the complex logarithm The confusing point here is that the formula $1^x = 1$ is not part of the definition of complex exponentiation, although it is an immediate consequence of the definition of natural number exponentiation
- Why is $1 i$ equal to $-i$? - Mathematics Stack Exchange
There are multiple ways of writing out a given complex number, or a number in general Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example
- arithmetic - Formal proof for $(-1) \times (-1) = 1$ - Mathematics . . .
The Law of Signs $\rm\: (-x)(-y) = xy\:$ isn't normally assumed as an axiom Rather, it is derived as a consequence of more fundamental Ring axioms $ $ [esp the distributive law $\rm\,x(y+z) = xy + xz\,$], laws which abstract the common algebraic structure shared by familiar number systems
- Why is $1$ not a prime number? - Mathematics Stack Exchange
actually 1 was considered a prime number until the beginning of 20th century Unique factorization was a driving force beneath its changing of status, since it's formulation is quickier if 1 is not considered a prime; but I think that group theory was the other force
- Why is $1^{\\infty}$ considered to be an indeterminate form
The indeterminate forms are often abbreviated with stuff like "$1^\infty$" but that's not what they mean This "$1^\infty$" (in regards to indeterminate forms) actually means: when there is an expression that approaches 1 and then it is raised to the power of an expression that approaches infinity we can't determine what happens in that form
- Proof that $(AA^{-1}=I) \\Rightarrow (AA^{-1} = A^{-1}A)$
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- 有问题，就会有答案 - 知乎
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