Let $A,B$ be elements of $M_2(\mathbb{R})$. Give an example to show that $A+B$ can be invertible if $A,B$ are both non-invertible

Question

The goal for this problem is to show that even if two matrices $A$ and $B$ are non-invertible, $A+B$ can be invertible. I tried to show this using a proof, but I ended up actually proving that this isn't true. I thought that it might just not be true, but the question before it was to prove the contrapositive, if $A,B$ are both invertible, show, by example, that $A+B$ can be non-invertible. I was able to prove that one, but using the same method I hit the end of the road proving this problem. Here is a basic outline of my steps for the problem. Please show me where I went wrong.

1) $$A = \begin{bmatrix} e & f\\ g & h \end{bmatrix}$$ and $$B = \begin{bmatrix} i & j\\ k & l \end{bmatrix}$$

2) $A+B=C$, so $$C = \begin{bmatrix} e+i & f+j\\ g+k & h+l \end{bmatrix}$$

After this, I say that the determinant of $C$ must not be 0 if it is to be invertible. If $A,B$ are non-invertible, then their determinants must be 0. Therefore,

det($A$): $eh-gf = 0$. Let $e=g=h=1$
$(1)(1) - (1)f = 0$, so $f$ must also be 1.

Therefore, $$A = \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}$$

I then did the same thing for $B$.

det($B$): $il - kj = 0$. If $l=1$, $i=2$, and $j=1$, then $k$ must be 2 for $B$ to be non-invertible. However, now that $$B = \begin{bmatrix} 2 & 1\\ 2 & 1 \end{bmatrix}$$

$A+B=C$, where $$C = \begin{bmatrix} 3 & 2\\ 3 & 2 \end{bmatrix}$$ Which is non-invertible as well.

Where did I go wrong?

Answer 1

Let $A=\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right)$. Each is non-invertible (with determinant $0$), but their sum is the identity and is invertible.

Answer 2

You tried to construct an example. The example you built didn't work. But there are plenty of others that will. For $A$, top left $1$, everybody else $0$. For $B$, bottom right $1$, all else $0$.

We can start with $A$ all $1$'s like you did. Then for $B$ many things will work, like $1$ down the main diagonal, and $-1$'s at the other corners. Other choices will be fine too, but if we choose very small integers, we may get unlucky.