TITLE: Showing $\ker(\alpha^h)=\ker(\alpha)$ for all $h\geq 1$, if $\alpha$ is self-adjoint
QUESTION [1 upvotes]: Let $ V$ be an inner product space and let $\alpha\in \operatorname{End}(V )$ be selfadjoint. Show that $\ker(\alpha) = \ker(α^h)$ for all $h \geq 1$.

Let $v\in \ker(\alpha)$, then $\alpha(v)=0$, so for any integer $h\geq1$, we have $\alpha^h(v)=\alpha^{h-1}(\alpha(v))=0$, which implies $v\in \ker(\alpha^h)$.
For the other part, I guess, it would be better to prove by indution on $h$.
The assumption is true for $h=2$.
Let $v\in \ker(\alpha^ 2)$, hence $\langle \alpha^2(v),v\rangle=\langle \alpha(v),\alpha(v)\rangle=0$ and this implies that $\alpha(v)=0$. Thus $v\in \ker(\alpha)$.
For the $h>2$ I don't know how to solve it and I also would like to have a prove with using the idea of endomorphism rather than matrices.

REPLY [1 votes]: We assume the assumption is true for all integers less than $h$.
Let $v\in ker (\alpha^h)$.
$$\alpha^{2(h-1)}(v)=\alpha^{h-2}(\alpha^h(v))=0\implies\langle\alpha^{2(h-1)}(v),v\rangle=\langle\alpha^{h-1}(v),\alpha^{h-1}(v)\rangle.$$
Thus, $\alpha^{h-1}(v)=0$ which implies $v\in ker(\alpha^{h-1})$. By the induction we have $v\in ker(\alpha).$


TITLE: dimension of graph
QUESTION [2 upvotes]: I have a really rudimentary question hope to clear out:
If $f: V \rightarrow V$, where $f$ is a linear map and $V$ is a vector space. Then what is the dimension of $\operatorname{graph}(f)$? To my understanding, $\operatorname{graph}(f)$ is $V \times V$, but with $\dim V$. Is this correct?

REPLY [2 votes]: As I said this question suppose to be a basic one, so user15464 answered it. Though other comments provide valuable resource and insight.
graph($f$) is the subset of $V \times V$ given by {$(v,f(v)):v\in V$}. It is homeomorphic to $V$ via projection onto the first coordinate.
For dimensions of manifolds, the dimension of the graph is the same as the dimension of $V$, since graph($f$) and $V$ are diffeomorphic.


TITLE: radius of convergence for $\sum \limits_{n=0}^\infty(-1)^n9^nx^{2n}$
QUESTION [1 upvotes]: I'm trying to find the radius of convergence for $\sum \limits_{n=0}^\infty(-1)^n9^nx^{2n}$. And I'm honestly new to this topic. So it would be cool if someone could tell me if my steps do make sense.
By using the root test I get:
$$\lim_{n \to \infty} \sqrt[n] {|(-1)^n9^nx^{2n}|} $$
$$=\lim_{n \to \infty} \sqrt[n] {|(-1)|^n} \cdot 9x^2$$
$$=\lim_{n \to \infty} 9x^2 = \infty$$
$$\Rightarrow R= 0$$

REPLY [3 votes]: The last line isn't correct: you should get
$$ \lim_{n\to\infty}[9^nx^{2n}]^{\frac{1}{n}}=\lim_{n\to\infty}9x^2=9x^2 $$
If this limit is less than $1$ then the series converges absolutely, while if the limit is greater than $1$ then the series diverges. Hence we need
$$ 9x^2<1 $$
or $|x|<\frac{1}{3}$, so $R=\frac{1}{3}$.


TITLE: Why is $y^3-5y^2+6y-3=0$ if $x^3+2x^2+3x+3=0$ where $ y = α/(α+1) $?
QUESTION [0 upvotes]: The beginning of a solution for a question in my book which is to find the value of $\sum α/(α+1)$ where α1, α2 and α3 are roots of $x^3+2x^2+3x+3=0$. For this in the solution they have said to take $ y = α/(α+1) $, and that therefore $y^3-5y^2+6y-3=0$. Where did this come from? How did they get this result? I tried substituting $ α = y/(1-y) $ but that just made it more complicated.

REPLY [1 votes]: Putting $x=\dfrac y{1-y}$ in $$x^3+2x^2+3x+3=0$$
$$y^3+2(1-y)y^2+3y(1-y)^2+3(1-y)^3=0$$
$$\implies y^3(-3-3-1+1)+y^2(2-6+9)+y(\cdots)+3=0$$