Diagonalization of a strange transformation

Question

Let be $V$ a vector space on $\mathbb C$ and $\dim V=4$ and let be $f \in \operatorname{End}(V)$ such that $\operatorname{Im}(f^2+a \cdot \operatorname{id}) \subset \ker(f+id)$ where $a=\det f$, $a\in\mathbb R$ and $a\ge0$.

1. Prove that for $a \ne 0$, $f$ is diagonalizable
2. Find for $a=2$ all the possible characteristic polynomials of $f$

Answer 1

Saying $\def\Im{\operatorname{Im}}\def\id{\operatorname{Id}}\Im(f^2+a\id) \subset \ker(f+\id)$ is equivalent to $(f+\id)\circ(f^2+a\id)=0$. So we are given

1. $(f+\id)\circ(f^2+a\id)=0$,
2. $a\in\Bbb R_{\geq0}$, and for question 1 even $a\in\Bbb R_{>0}$
3. $a=\det(f)$,
4. $f$ acts on a $4$-dimensional space, so the characteristic polynomial $\chi_f$ has $\deg(\chi_f)=4$.

Conditions 1,2 alone ensure that $f$ is diagonalisable over$~\Bbb C$, since the polynomial that annihilates$~f$ is $\def\i{\mathbf i}(X+1)(X^2+a)=(X+1)(X-\i\sqrt a)(X+\i\sqrt a)$, whose factors have distinct roots if $a>0$. It is a theorem that any linear operator annihilated by a polynomial that factors into distinct monic factors of degree$~1$ is diagonalisable. But if you don't know that theorem, here is how one can reason elementarily for this case. The kernel and image of $f^2+a\id$ are complementary subspaces: by rank-nullity their dimensions are complementary, and if $v\in\ker(f^2+a\id)\cap\Im(f^2+a\id)$ then $v\in \Im(f^2+a\id)$ and hypothesis 1. together give $f(v)=-v$, and then $v\in\ker(f^2+a\id)$ gives $(-1)^2v+av=(1+a)v=0$; by hypothesis 2. one has $a\neq-1$, so $v=0$ is forced (note that we used that the root $-1$ of $X+1$ is not a root of $X^2+a$). So $V=\ker(f^2+a\id)\oplus\Im(f^2+a\id)$, and the second summand is contained in the eigenspace of$~f$ for$~{-}1$ so it suffices that the restriction $\tilde f$ of$~f$ to the first summand is diagonalisable. But $(\tilde f+\i\sqrt a\id)\circ(\tilde f-\i\sqrt a\id)=0$, this subspace decomposes similarly as direct sum of kernel and image of $\tilde f-\i\sqrt a$ by the same kind of reasoning, using that $\i\sqrt a\neq-\i\sqrt a$ since $a\neq0$. Thus the sum of the eigenspaces for $-1$, $\i\sqrt a$ and $-\i\sqrt a$ is all of$~V$.

For question 2, you need to use the hypotheses 3., 4. We have seen that the only possible eigenvalues are $-1$, $\i\sqrt a$ and $-\i\sqrt a$, but hypothesis 3. requires their product, taken with multiplicities, to be $a$, and by hypothesis 4. the total multiplicity is $4$. For $a=2$ this can only be achieved if one has multiplicity $2$ for the eigenvalue $-1$, and multiplicity $1$ for the other two: $$\chi_f=(X+1)^2(X-\i\sqrt2)(X+\i\sqrt 2)=(X^2+2X+1)(X^2+2)=X^4+2X^3+3X^2+4X+2. $$ In general (but still with hypothesis 2) other combinations of multiplicities may work: having only an eigenvalue $-1$ with multiplicity$~4$ (giving $\det(f)=(-1)^4=1$, so it requires $a=1$), or having no eigenvalue $-1$ but rather both $\i\sqrt a$ and $-\i\sqrt a$ each with multiplicity$~2$ (giving $\det(f)=a^2$, so it requires $a^2=a$, that is $a=1$ (again!) or $a=0$). And remember that in the $a=0$ case, $f$ need not be diagonalisable (any $f$ with $f^2=0$ will do there).

Answer 2

1.$$ 0 = (f+I)(f^2 + a I)(x) $$now if $a\neq 0$:

$$ (X+1)(X^2 + a) $$has simle roots on $\Bbb C$, and then $f$ is diagonalizable.

1. The eigenvalues are roots of the polynomial, that is if $a=2$: $$ -1; \pm \sqrt{2} $$ so the possible characteristic polynomials are $$ (-X-1)^a (-X+\sqrt{2})^b (-X-\sqrt{2})^c;\\ a + b +c = 4 $$