# Thread: real symmetric similar matrix

1. ## real symmetric similar matrix

I have a matrix A which is positive definite and having all real elements. A can have multiple eigenvalues but it is not defective.
I want to get a similar matrix B (i.e., B has the same eigenvalues as A) which is symmetric and real. So precisely I want to construct a nonsingular matrix X such that B = inv(X) A X.
So it will be extremely helpful if anyone can tell me how to get such X or at least give me some reference.

Alternatively:
A method is available to get a symmetric similar matrix for A from another nonsingular symmetric matrix Y if it satisfies A Y = Y A_tr. But such similar matrix obtained from Y will only be real if Y is positive definite. So alternatively, any help in getting a positive definite Y for given A will be extremely useful.

2. Originally Posted by sandipdas76
i have a matrix a which is positive definite and having all real elements. A can have multiple eigenvalues but it is not defective.
I want to get a similar matrix b (i.e., b has the same eigenvalues as a) which is symmetric and real. So precisely i want to construct a nonsingular matrix x such that b = inv(x) a x.
So it will be extremely helpful if anyone can tell me how to get such x or at least give me some reference.

Alternatively:
A method is available to get a symmetric similar matrix for a from another nonsingular symmetric matrix y if it satisfies a y = y a_tr. But such similar matrix obtained from y will only be real if y is positive definite. So alternatively, any help in getting a positive definite y for given a will be extremely useful.

If we assume A is a symmetric and real, then B will be real; however, not sure if that implies B will be symmetric. If it does imply that B is symmetric, then this is how I would show it.

$A=A^T$
$B=X^{-1}AX$

$B^T=(X_1^{-1}AX_1)^T=X_1^TA^T(X_1^{-1})^T=((X_1^T)^{-1})^{-1}A^T(X_1^{-1})^T$
$X=(X_1^T)^{-1}$
$B^T=X^{-1}A^TX$

3. Thanks for your reply. The main problem is that the matrix $A$ is not symmetric. If it is so then there is no need to get any symmetrizer for that. I can work with $A$ itself.