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Apr
24
comment Find the bases for the eigenspaces of the matrix.
Since the matrix you got was the zero matrix, every vector is an eigenvector for eigenvalue $1$. So, you can just choose the standard basis.
Apr
21
answered Mapping a circle to a point on a sphere
Apr
12
answered Linear Algebra Help: Change of Basis Matrix
Apr
12
revised Linear Algebra Help: Change of Basis Matrix
added 35 characters in body
Apr
2
comment Why is it called the *Inverse* Galois Problem?
Extensions of rational numbers will produce groups via the Galois Correspondence. We want to know whether the reverse is true: Every group produces an extension.
Mar
31
revised An exponential equation
deleted 55 characters in body
Mar
31
comment What is the difference between the three types of logarithms?
I would think $\ln$ and $\log$ are the same and that $\operatorname{Log}$ refers to the principal logarithm as listed here en.wikipedia.org/wiki/Complex_logarithm under "Definition of Principal Value".
Mar
14
revised If a matrix $A^2$ is invertible, is $A^3$ invertible?
added 4 characters in body
Mar
14
answered If a matrix $A^2$ is invertible, is $A^3$ invertible?
Mar
4
answered Show that if $\lim_{k\to\infty} x_k= -\infty$, then $\lim_{k\to\infty} \frac{1}{x_k} = 0$
Feb
27
comment What are the roots of this function with absolute values?
If $x\neq 0$, then $|x| > 0$ and $x^2 > 0$.
Feb
26
asked How Does This Fourier Grapher Work?
Feb
16
revised Can $\langle x,x \rangle$ be less than zero on an inner product space?
added 6 characters in body
Feb
12
comment If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
@user236182 No, it should not. The OP has shown that we need only consider the $f(x)$ in my answer.
Feb
12
comment If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
@user19405892 I have added a proof that there is only one root. But, as I said, it doesn't stay at the precalculus level, which is what you tagged the question.
Feb
12
revised If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
added 573 characters in body
Feb
12
comment If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
@Hetebrij I realized that as well. But, at the precalculus level, I see no rigorous way of showing it is the only one.
Feb
12
comment If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
@Kf-Sansoo The OP doesn't mention that anywhere in their post.
Feb
12
answered If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
Feb
12
comment Show that $\partial A$ is always a closed set
The interval $(1,3]$ is neither open nor closed.