I found the following statement in a book:
For matrices $A$ and $B$ both symmetric and positive definite, if the eigenvalues of: $$A^{1/2}B^{-1}A^{1/2},$$ are all equal to 1, than $A$ is identical to $B$.
How can I prove such statement?
Eigenvalues of "$A^{1/2}B^{-1}A^{1/2}$" equal to 1
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linear-algebra
eigenvalues-eigenvectors
positive-definite
1 Answers
2
First, the matrix $A^{1/2}B^{-1}A^{1/2}$ is symmetric for obvious reasons. All its eigenvalues are equal to $1$, and the only symmetric matrix with such a property is the identity matrix, hence $$A^{1/2}B^{-1}A^{1/2}=I.$$ After that multiply this identity by $A^{-1/2}$ from the left and from the right, which leads to $$B^{-1}=A^{-1}$$ and $$B=A.$$