# A question of non-singularity

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.

• I edited your post a little to make the $\LaTeX$ work etc. Your can see how I did it by clicking the "edit" button. Cheers! – Robert Lewis Oct 5 '18 at 18:29

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.

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).

• @StubbornAtom: Oops! Typo; corrected. Thanks. Cheers! – Robert Lewis Oct 5 '18 at 19:10
• @StubbornAtom: damn! another blunder! Now how does it look? That's what I get for getting on MSE before coffee! Cheers! – Robert Lewis Oct 5 '18 at 19:23
• How do you exclude the other cases? – egreg Oct 5 '18 at 20:48
• @egreg I didn't. I guess I will have to cite your answer! (+1) – Robert Lewis Oct 5 '18 at 22:05

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 $$B+2I=0$$?

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.