Question #194356

 Let A be any n x n matrix and let P be an n x n orthogonal matrix. Prove that jAj D jP


APj:


1
Expert's answer
2021-05-19T17:25:13-0400

A is n×nn\times n symmetric matrix.

its all eigen values are greater than 0.


Characteristics polynomial is det(AλIn)=0det(A-\lambda I_n)=0


λ\lambda are the eigenvalues of A.

As P is orthogonal matrix so PT=P1P^T=P^{-1}


Characteristics polynomial for PTAPP^TAP is given as-


det(PTAPλIn)=det(P1APλP1P)=detP1(AλIn)P=det(P1)det(P)det(AλIn)=det(P1P)det(AλIn)=det(AλIn)det(P^TAP-\lambda I_n) \\ =det(P^{-1}AP-\lambda P^{-1}P) \\ =det{P^{-1}(A-\lambda I_n)P} \\ =det(P^{-1})det (P)det(A-\lambda I_n) \\ =det(P^{-1}P)det(A-\lambda I_n) \\ =det(A-\lambda I_n)



(PTAP)T=(PA)T(PT)T=PTATP=PTAP,asAT=A(P^TAP)^T=(PA)^T(P^T)^T=P^TA^TP=P^TAP, as A^T=A


So PTAPP^TAP is symmetric matrix.


From characteristics equation, eigenvalues of PTAPP^TAP is the same as eigenvalues of A.


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