# How to prove there are $n$ independent eigenvectors

A given $$n\times n$$ matrix $$B$$ of real entries obeys $$B^{2}=-I$$.

I know that $$B$$ has only 2 eigenvalues $$i,-i$$ and that $$n$$ has to be even. Therefore each eigenvalue must have $$n/2$$ multiplicity. How do I prove that there are $$n$$ linearly independent eigenvectors?

• Observe that $x^2+1$ is the characteristics polynomial as well as minimal polynomial of $B$ which has distinct linear factors and hence $B$ is diagonalizable and therefore it has $n$ linearly independent eigen vectors. – little o Feb 28 '19 at 15:49
• Just to clarify: $x^2 +1$ is not the characteristic polynomial of $B$ unless $n=2$. But it is the minimal polynomial. – cs47511 Feb 28 '19 at 16:05
• @Dbchatto67 Any way of showing that there are $n$ linearly independent eigenvectors without invoking that $B$ is diagonalizable? – user2175783 Feb 28 '19 at 16:06
• @csprun you are correct. – little o Feb 28 '19 at 16:07
• Those two conditions are equivalent, and @Dbchatto67 hasn't invoked it, he has proved it (albeit concisely). One can prove that if a matrix $B$ satisfies a polynomial with no repeated roots, e.g. $x^2+1$ over $\mathbb{C}$, then it's diagnalizable (in an algebraic closure of the ground field). Again, you can prove this easily with JNF. – cs47511 Feb 28 '19 at 16:07

This is essentially the same as what @Dbchatto67 says in the comments, but if you know about Jordan normal form (very much worth learning if not), then you can solve this easily by thinking in JNF. A $$k\times k$$ Jordan block (over $$\mathbb{C}$$) $$J = \begin{pmatrix} \lambda & 1 & 0 & \dots & 0 & 0 \\ 0 & \lambda & 1 & 0 & \dots & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \dots & \lambda \end{pmatrix}$$ does not satisfy $$J^2 = -I$$ unless $$\lambda = \pm i$$ (as you already saw) and $$k=1$$. So the JNF of $$B$$ over $$\mathbb{C}$$ is diagonal and there are $$n$$ linearly independent eigenvectors over $$\mathbb{C}$$.