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Assume that $A$ and $B$ are invertible matrices, is $A+B$ (given that $A+B\neq0$) an invertible matrix too?

And if so how do I prove it?

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3 Answers

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No, not in general.

$A+B=0$ is not the only way to reduce the rank of the sum, you can arrange to have a matrix full of $1$ for instance:

$A=\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}\quad B=\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}\quad A+B=\begin{bmatrix} 1 & 1\\ 1 & 1\end{bmatrix}$

$\det(A)=1$ and $\det(B)=-1$ while $\det(A+B)=0$.

Or a less extreme example, with columns that are linearly dependant

$A=\begin{bmatrix} 1 & 3\\ 2 & 4\end{bmatrix}\quad B=\begin{bmatrix} 2 & 3\\ 1 & 2\end{bmatrix}\quad A+B=\begin{bmatrix} 3 & 6\\ 3 & 6\end{bmatrix}$

$\det(A)=-2$ and $\det(B)=1$ while $\det(A+B)=0$.

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On a more theoretical point of view, notice that for a matrix $A$ invertible and for $B=-\lambda I$ (invertible too) then $A+B=A-\lambda I$ is not always invertible.

Indeed when $\lambda$ is an eigenvalue of $A$ then $\det(A-\lambda I)=0$ meaning exactly that $A+B$ is not invertible.

So it is not just "bad luck" that the sum of two matrices is not always invertible, this is in fact a fundamental aspect of linear algebra.

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Given $A$ an invertible matrix, take $B=-A$, you have the null matrix that is not invertible

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Update the question has been updated with the extra constraint that $A+B\neq 0$.

Here is a new counter-example, take:

$$ A=\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right ),\ B=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right ) $$

You have $det(A)=1$ and $det(B)=-1$, hence both are invertibles, however:

$$ A+B=\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right ) $$

is obviously not invertible.

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Note: we can notice that for positive-definite matrices the result is true: if $A$ and $B$ are positive-definite matrices then $A+B$ is also a positive-definite matrix, hence invertible.

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Let $I$ be the identity matrix. $I$ and $-I$ are invertible, but $I + (-I)$, or the zero matrix, is not.

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