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Let $A$ be a symmetric invertible $n \times n$ matrix, and $B$ an antisymmetric $n \times n$ matrix. Under what conditions is $A+B$ an invertible matrix? In particular, if $A$ is positive definite, is $A+B$ invertible?

This isn't homework, I am just curious. Assume all matrices have entries in $\mathbb{R}$.

Edit to include context:

This question comes from a question that popped up in my research on string theory. One is interested in (pseudo)-Riemannian manifolds equipped with a two-form gauge field, modelling a background in which a closed string is moving. The metric, $g$, is a symmetric covariant 2-tensor, while the $b$-field is an antisymmetric covariant 2-tensor. The metric is non-degenerate and therefore invertible. Choosing local coordinates for the manifold, we can express the metric and $b$ field as $n \times n$ matrices, say $A$ and $B$, where $A$ is invertible. There is an operation on string backgrounds called T-duality which, in this simplified context, acts by inverting the matrix $E = A + B$, and so I am therefore interested in which scenarios this procedure works. I am mainly interested in the context where $A$ is real, invertible and positive definite (positive eigenvalues), corresponding to a Riemannian metric $g$, although I have tried to be a bit more general in the wording of the question.

Where to start: The main issue I have is that I don't really have any criteria for when the sum of two matrices is invertible. Certainly if the determinant is non-zero then I will be happy, but the determinant is not additive, so I don't know how to approach this. In two dimensions I can construct a counterexample whenever A has negative determinant, but the situations I really care about have $det(A)>0$. I would like to find a general criterion for when $A+B$ is invertible.

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    $\begingroup$ $$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 1& -1 \end{pmatrix} $$ $\endgroup$
    – Chappers
    May 3 '18 at 1:45
  • $\begingroup$ What makes you believe this might be the case? $\endgroup$
    – Asinomas
    May 3 '18 at 1:45
  • $\begingroup$ Whether it's homework or not, you're expected to contribute some effort when you ask a homework-style question. $\endgroup$
    – amWhy
    May 3 '18 at 17:35
  • $\begingroup$ Edited to include context and my approach. $\endgroup$
    – Mark B
    May 3 '18 at 23:17
  • $\begingroup$ If you are interested in the conditions under which $A+B$ is invertible, perhaps you should ask that question, instead? You might edit the title and body of the question to make that point. $\endgroup$
    – Xander Henderson
    May 3 '18 at 23:48
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Pick $B$ any anti-symmetric matrix which is not nilpotent, and $\lambda \neq 0$ an eigenvalue of $B$.

Set $$A=-\lambda I$$

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  • $\begingroup$ perfect solution, $+1$. $\endgroup$
    – Asinomas
    May 3 '18 at 1:46
  • $\begingroup$ Seconded. +2 ${}$ $\endgroup$
    – David Hill
    May 3 '18 at 2:21
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Nah. Take $$A = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \quad \mbox{and} \quad B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$$Then $A$ is invertible and $$A+B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$has determinant $\det(A+B) = 0$.

(Every matrix is the sum of a symmetric matrix and an anti-symmetric matrix. Take a non-invertible matrix, decompose it like that and see if the symmetric part is invertible.)

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  • $\begingroup$ This is precisely the context of where this question came from. I am interested in the situations in which the symmetric part ${is}$ invertible. In particular, I am looking for a solution which has $det(A) > 0$. $\endgroup$
    – Mark B
    May 3 '18 at 1:47
  • $\begingroup$ If $A$ is invertible, then the matrix determinant lemma says that $$\det\left(\frac{A+A^\top}{2}\right) = \frac{\det A}{2^n} \det({\rm Id} + A^\top A^{-1}).$$ $\endgroup$
    – Ivo Terek
    May 3 '18 at 1:52
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If $A$ with positive determinant then in the case $n=2$, $A+B$ will be invertible. For $n\ge 3$, you can adjust the $2\times 2$ examples given in the other answers by adding a piece $(-1,1,\ldots)$ on the diagonal to make $A$ $n\times n$ with positive determinant.

Maybe you are thinking about the following: if $A$ is positive definite then the real parts of the eigenvalues of $A+B$ are positive ( and so $\ne 0$). This is easy to show but requires to look at complex vectors. A bit more general, considering complex matrices:

If $A_1$, $A_2$ are hermitian then the real parts of the eigenvalues of $A_1+ i A_2$ are between the the smallest and the largest eigenvalue of $A_1$, and the imaginary parts of the eingenvalues of $A_1 + i A_2$ are between the smallest and the largest eigenvalue of $A_2$.

Note that if $B$ real and skew symmetric then $1/i B$ is hermitian. Therefore, the eigenvalues of $B$ are purely imaginary and come in pairs, $i b$, $-i b$ (and perhaps some zero eigenvalues). So in the real case $B$ has "no input" towards invertibility of $A+B$. But a definite $A$ will guarantee invertibility of $A+B$.

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