Prove that at least one of the matrices is invertible

Question

The question is to show that if there is a matriz $C \in \mathbb{M}_3 (\mathbb{C})$, then at least one of $C + I_3$, $C + 2I_3$, $C + 3I_3$, $C + 4I_3$ is invertible.

My first approach was by eigenvalues. $C + I_3 = C - (-I_3)$ and $-1$ is the only eigenvalue of $-I_3$, so I tried to prove that $det(C + I_3 - \lambda I_3) = det(C - (-I_3) - \lambda I_3) = det(C- (-1 + \lambda)I_3)$ was equal to $0$ or not.

So, if $\lambda = 1$, then $det(C - (-1 + \lambda)I_3) = det(C)$, but we don't know anything about $C$ to know if $det(C) = 0$ or not. If $\lambda \neq 1$, then $det(C - (-1 + \lambda)I_3) = det(C - \gamma I_3)$, $\gamma = -1 + \lambda$, which inplies that $C - \gamma I_3 \neq 0$, so that $C + I_3$ would be invertible. Anyay, I didn't get to any conclusion. The idea as to do this for each of the four matrices.

My second approach was by contradiction. So, I supposed that any of these matrices was invertible. Then, $det(C + I_3) = det(C + 2I_3) = det(C + 3I_3) = det(C + 4I_3) = 0$. But, this implies that $C + I_3 = \cdots = C + 4I_3$, which is the same as saying that $C = -I_3 = \cdots = -4I_3$, which is not true. So, we would have that at least one of the four matrices is invertible. Anyway, I don't think my proof is correct.

Any suggestions? Thanks.

Answer 1

I am assuming that the correct statement is the next one:

Given any matrix $C\in\Bbb M_3(\Bbb C)$, at least one of the matrices $C+kI_3$ ($k=1,2,3,4$) should be invertible.

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By way of contradiction, assume that there is some matrix $C\in\Bbb M_3(\Bbb C)$ such that $C+kI_3$ is not invertible for $k=1,2,3,4$.

Then $\det(C+k I_3)=0$ $k=1,2,3,4$ and this would imply that $M$ has 4 differents eigenvalues (say, $\lambda=-1,-2,-3,-4$). BUt this is impossible since $M$ is a 3x3 matrix.

So at least one of these matrices would have nonzero determinant.

Answer 2

Hint : $$\det(C+XI_n)$$ is a nonzero polynomial of degree $3$, hence it cannot have four distinct roots.