Bases de una matriz

Question

Suppose that V is a vector space over $ \mathbb{R} $ of dimension $dim_{ \mathbb{R} } V = 2$ and $J\in End_{ \mathbb{R} }(V )$ satisfies $J^2=-1$. Show that there exists a basis of V with respect to which the matrix of J is equal to $\begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix}$

Answer 1

Hint: Take $v_1 \in V$ with $v_1\ne0$. Let $v_2=Jv_1$. Then prove that $v_1$ and $v_2$ are linearly independent.

Indeed:

If $v_2=\lambda v_1$, then $Jv_1=\lambda v_1$ and so $-v_1 = J^2v_1=\lambda Jv_1=\lambda^2 v_1$. Therefore, $\lambda^2=-1$, which cannot happen for $\lambda \in \mathbb R$.

Answer 2

Note that $P(x) =x^2+1$ satisfies $P(J)=0$. Thus, the characteristic polynomial divides it and because both are of degree 2 and monics, $P$ is the characteristic of $J$.

Think $J$ over $\Bbb{C}$, and because P has simple roots, $J$ is diagonalixable with autovalues $\pm i$. Then, exist a basis $ B$ such that $|f|_B=diag(i,-i)$

Note that the matrix M of exercise have the same characteristic and exist $B'$ such that $M=C(B',E)diag(i,-i)C(E,B')$ where $E$ is the canonical basis.

Now can you finish?