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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?

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You never appeared to use the equation for $\mbox{det}(C) = 0$ when you chose the elements of $A,B$. When trying to prove something, sometimes it's best to write down full equations and go through everything in the most generality to see what's going on. This isn't one of those cases. You just need one counterexample. You need $A, B$ to be invertible, and $C$ to not be invertible. What's the most non-invertible matrix you can think of, and how might you get that from adding invertible matricies? Just play around with some concrete examples before writing out all the equations. – Zach L. May 30 '13 at 3:19
Your idea of what is a contrapositive is wrong, but beware that you are asserting that an implication does not hold. To prove "by contrapositive" that $A,B$ both singular does not imply $A+B$ singular, you could show that $A+B$ non-singular (i.e., invertible) does not imply that $A,B$ are both non-singular. Your "question before" amounts to "$A,B$ both non-singular (invertible) does not imply $A+B$ non-singular", which is quite something different. Maybe you meant "converse" rather than "contrapositive", but even using that more vague term here is questionable. – Marc van Leeuwen May 30 '13 at 4:42
up vote 10 down vote accepted

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.

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Thanks. I guess I tried to take a too complicated method. I should have started with $C$ being the Identity and 2 by 2, and then done some simple stuff for A and B. Thanks again. – EW - CodeMonkey May 30 '13 at 3:27

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.

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You win by 15 seconds. :-) – vadim123 May 30 '13 at 3:16
Thank you. When I refreshed your answer was first, so I accepted it. I should have taken a very simple approach. – EW - CodeMonkey May 30 '13 at 3:29
@EW-CodeMonkey: As a matter of principle you should accept the best answer, not the first correct one. You are of course entirely free to choose which criteria define "best" for you. – Marc van Leeuwen May 30 '13 at 4:49

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