Using the equation $ A^{-1} + B^{-1} = (A+B)^{-1} $ to find relations between det(A) and det(B)

Question

We are given a condition,

$$ A^{-1} + B^{-1} = (A+B)^{-1} $$

Further, $|A| =4$ and we are asked to find the value of |B|.

I tried to simplify the LHS $$ A^{-1} + B^{-1}= B^{-1}BA^{-1} + B^{-1}AA^{-1} $$ $$ A^{-1} + B^{-1}= B^{-1}(A+B)A^{-1} $$

That only lead me to get

$$ |A+B|^2=|A||B| $$

Am stuck after this. Any help would be appreciated! Thanks.

Answer 1

If $A,B$ have sizes $n$ with $n > 2$, then it is impossible to determine $|B|$.

If $A,B$ have real entries and $n$ is odd, then it is impossible for $A,B$ to satisfy the given condition.

If $n = 2$ and $A,B$ have real entries, then $|B| = |A|$ must hold.

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We can write $$ A^{-1} + B^{-1} = (A + B)^{-1} \iff\\ A(A^{-1} + B^{-1}) = A(A + B)^{-1} \iff\\ AA^{-1} + AB^{-1} = ((A + B)A^{-1})^{-1} \iff\\ I + AB^{-1} = (I + BA^{-1})^{-1}. $$ Writing $M = BA^{-1}$ (or equivalently $B = MA$) lets us rewrite this as $$ I + M^{-1} = (I + M)^{-1} \iff\\ (I + M)(I + M^{-1}) = I \iff\\ M + 2I + M^{-1} = I \iff\\ M + I + M^{-1} = 0 \iff\\ M^2 + M + I = 0. $$ In other words: invertible matrices $A,B$ will satisfy the given equation if and only if $B = MA$ for some (invertible) matrix $M$ with $M^2 + M + I = 0$. $M$ will satisfy this equation if and only if it is diagonalizable and its eigenvalues satisfy $\lambda^2 + \lambda + 1 = 0$. In other words, $M$ will satisfy the equation iff its eigenvalues are all equal to either of the two possibilities $\lambda = -\frac 12 \pm i\frac{\sqrt{3}}2$.

For a $2 \times 2$ real matrix, this implies that $|M| = 1$, so that $|B| = |MA| = |A|\cdot |M| = |A|$.

Answer 2

Hi and welcome to MSE.

Your condition is not sufficient to determine an exact value of $\det(B)$. For, take the vector space $\mathbb C$. A linear application $T_c:\mathbb C \rightarrow \mathbb C$ is just a multiplication for an element $c$ of $\mathbb C$ and the determinant is exactly $c$. So the matrix that represent $T_c$ is just the $1\times 1$ matrix $(c)$.

Take now the linear application $A=T_4$ (that is an application with determinant equal to $4$). Let $B=T_b$; thanks to your condition we have: \begin{gather} A^{-1}+B^{-1} = (A+B)^{-1}\\ T_4^{-1}+T_b^{-1} = (T_4+T_b)^{-1}\\ \end{gather}

Using now the fact that $T_c^{-1}=T_{\frac{1}{c}}$ and $T_c + T_d = T_{c+d}$ we obtain the following equation: \begin{equation} \frac{1}{4} + \frac{1}{b} = \frac{1}{4+b} \end{equation} that has two solutions: \begin{gather} b_1 = -2 + 2i\sqrt 3\\ b_2 = -2 - 2i\sqrt 3 \end{gather} Then, you have two linear application $B_1=T_{b_1}$ and $B_2 = T_{b_2}$ that satisfie the initial condition but they have different determinant $\det(B_1)=b_1\neq b_2=\det (B_2)$. So the initial conditions are not sufficient to determine an exact value of $\det(B)$.

Answer 3

From $(A+B)(A^{-1}+B^{-1})=I$ you get $$ I+AB^{-1}+BA^{-1}+I=I \tag{1} $$ and therefore $$ I+AB^{-1}+BA^{-1}=0 \tag{2} $$ Multiply $(2)$ on the right by $A$: $$ A+AB^{-1}A+B=0 \tag{3} $$ Multiply $(2)$ on the right by $B$: $$ B+A+BA^{-1}B=0 \tag{4} $$ Therefore, comparing $(3)$ and $(4)$, $$ AB^{-1}A=BA^{-1}B\tag{5} $$ If $a=\det A$ and $b=\det B$, we get from $(5)$, using Binet's theorem, $$ a^2b^{-1}=a^{-1}b^2 \tag{6} $$ Therefore $b^3=a^3$ and, assuming matrices with real coefficients, $b=a$.

Notes.

1. This answer doesn't touch the problem of existence of the given matrices. Other answers give conditions about this.

2. If the matrices are allowed to have complex entries, the only possible conclusion is that $b/a$ is a cube root of unity.