If $X^2+I$ is not invertible, it is $0$ Let $X \in M(\mathbb{R},2)$. If $X^2+I$ is not invertible, it is $0$.
I could prove it by dirty calculation, but it takes too much time.How can you prove it easily?
 A: $0$ is an eigenvalue of $X^2+I$, hence there is an eigenvalue $ \mu$ of $X$ such that $0=\mu^2+1$. This gives $\mu=i$ or $ \mu=-i$. Since the char. polynomial $p$ of $X$ has real coefficients, we get $p(x)=(x-i)(x+i)=x^2+1$. By Cayley-Hamilton: $p(X)=0$.
A: Using Jordan form,$P^{-1}(X^2+I)P$ has eigenvalue $i$ or $-i$. But $X$ is real, so $X$ has  both $i$ and $-i$. Then $X^2 + I$ is $0$
A: Are these the dirty calculations? They seem clear to me...
Since $X^2+I$ is not invertible, there is $v \in V$, $v \ne 0$, such that $(X^2+I)v=0$.
Let $w=Xv$. Then $Xw=X^2v=-v$ and so $X^2w=-Xv=-w$, that is $(X^2+I)w=0$.
Now $v$ and $w$ are linearly independent and so form a basis of $V$. Indeed, $\alpha v + \beta w = 0$ implies $0=X(\alpha v + \beta w)=\alpha w - \beta v$. Therefore $(\alpha^2+\beta^2)v=0$ and so $\alpha=\beta=0$.
Thus, $X^2+I=0$ is equal to $0$ on a basis of $V$ and so $X^2+I=0$.
A: This can also be proved by using only basic matrix algebra, without mentioning complex numbers, eigenvalues or vector space.
For $2\times2$ matrices, one can verify the following identities directly:
\begin{align}
&\det(Y+I_2) = \det(Y) + \operatorname{tr}(Y) + 1,\tag{1}\\
&\operatorname{tr}(X^2) = \operatorname{tr}(X)^2 - 2\det(X),\tag{2}\\
&X^2-\operatorname{tr}(X)X+\det(X)I_2=0.\tag{3}
\end{align}
By assumption, $X^2+I_2$ is not invertible. Put $Y=X^2$, the LHS of $(1)$ becomes zero. Substitute $(2)$ into $(1)$, we get $0 = \det(X)^2 + \operatorname{tr}(X)^2 - 2\det(X) + 1$. That is, $\left(\det(X)-1\right)^2 + \operatorname{tr}(X)^2 = 0$. As the determinant and trace are real, we get $\det(X)=1$ and $\operatorname{tr}(X)=0$. Now the conclusion follows from $(3)$.
