# Finding a base of a matrix.

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.

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.
• 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? – user63181 Jun 22 '14 at 17:11