A question of non-singularity

Question

Let $A$ and $B$ be matrices such that $B^2+ AB + 2I = 0$, where I denotes the identity matrix. Which of the following matrices must be nonsingular?

(A) $A + 2I$

(B) $B$

(C) $B + 2I$

(D) $A$

I tried using a few tricks assuming each option to be nonsingular and then coming to the given form but to no avail. Any hint is appreciated.

Answer 1

Since $$\det (B+A)\cdot \det (B) = \det (B^2+AB) = \det (-2I) =-2$$ we see that $\det (B)\ne 0$ so $B$ is invertibile.

Answer 2

We have

$B^2 + AB + 2I = 0, \tag 1$

so

$(B + A)B = B^2 + AB = -2I, \tag 2$

or

$\left ( -\dfrac{1}{2}(B + A) \right )B = I, \tag 3$

that is,

$B^{-1} = -\dfrac{1}{2}(B + A), \tag 4$

and the correct answer is (B).

Answer 3

I'm assuming you're over the real numbers.

Suppose $B$ is singular and take $v\ne0$ such that $Bv=0$; then $$ 0=(B^2+AB+2I)v=B^2v+ABv+2Iv=2v, $$ a contradiction. Therefore $B$ is nonsingular.

In order to exclude the other cases, let's see whether $A$, $A+2I$ or $B+2I$ can be the zero matrix.

The matrix $A$ can be the zero matrix, because $$ \begin{bmatrix} 0 & 2 \\ -1 & 0 \end{bmatrix}^2=-2I $$

Also $A+2I$ can be the zero matrix: the identity to satisfy would be $$ B^2-2B+2I=0 $$ and the matrix \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} satisfies it. I looked for a matrix with trace $2$ and determinant $2$, so Hamilton-Cayley solves the problem.

Can $B+2I=0$?

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Just for completeness, the same computations can be performed over any field, with the same result, provided the field doesn't have characteristic $2$. Indeed, the initial contradiction we found is definitely not such when $2=0$.

In this case the identity becomes $B^2+AB=0$ and we see that $B$ can as well be the zero matrix, with $A$ any matrix at all. If $B$ is invertible, then $A=B$.

There is no case distinction for $A+2I=A$ and $B+2I=B$. Thus in the case of characteristic $2$, none of the four conditions is forced.