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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.