I'm struggling with a question where I know an application but I need to find the base of a given matrix.

set $u : E \to E$ with E an R-vector space in n dimension. (sorry for my english)

it is given that $$u²+4Id=0$$

a few questions before this one, I had to prove that u didn't have any eigenvalue and wasn't bijective.I could prove it and hope I could use $ker(u-\lambda Id)\neq0$ and $det(u)\neq0$

then, here is the Question : $n = 2$ I'm asked to demonstrate if there is an existing base of E where the matrix of $u$ is:

$$A=\begin{pmatrix} 0&&-4 \\ 1&&0 \\ \end{pmatrix}$$

so when I saw that question I thought I had to prove that this matrix is similar to another matrix of $u$ in a different base so I tried :

set $\forall x \in E$, let's take $(u(x),x)$ (I don't know what other base I can take)

$\forall x \in E$, $\lambda_1 u(x)+\lambda_2 x =0$ we know that $ker(u-\lambda Id)\neq0$ which means that $\forall x \in E , u(x)\neq \lambda x$ if $x\neq0$

but the problem is that I don't know if I'm allowed to suppose my $x\neq 0$ to say my family is linearly independant. Besides, even though it may be a base of E, I'm not convinced of having the base of my matrix given upthere.

thank you in advance !


Notice that $u\ne0$ so $\ker u\ne \Bbb R^2$ so let $x\not\in\ker u$ hence $u(x)\ne0$ and then the set $(u(x),x)$ is linearly independent. In fact, if we have $\alpha,\beta\in\Bbb R$ such that

$$\alpha u(x)+\beta x=0\tag1$$ then we apply $u$ to $(1)$ to get $$-4\alpha x+\beta u(x)=0\tag2$$ and from $(1)$ and $(2)$ we find $\beta^2+4\alpha^2=0$ and then $\alpha=\beta=0$. Conclude.

  • $\begingroup$ thank you very much ! I also wanted to know how can my base is related to my matrix ? Should I just answer by showing my chosen base is linearly independant ? $\endgroup$ – Brocolus Jun 22 '14 at 17:08
  • $\begingroup$ Isn't clear for you that the matrix of $u$ relative to the basis $(x,u(x))$ takes the form of $A$ given in your question? $\endgroup$ – user63181 Jun 22 '14 at 17:11
  • $\begingroup$ well I know it should be clearer but I'm not very sure. I see here u(u(x))=-4x (because of u²+4Id=0) and u(x)=u(x) is it a way of seeing it ? $\endgroup$ – Brocolus Jun 22 '14 at 17:19
  • $\begingroup$ Yes that's correct. $\endgroup$ – user63181 Jun 22 '14 at 17:21
  • $\begingroup$ Alright thank you very much ! $\endgroup$ – Brocolus Jun 22 '14 at 17:22

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