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Find the eigenvalues and eigenvectors of the real symmetric matrix

M=(AHHB)

Show that the eigenvalues are real and the eigenvectors are perpendicular.

Short Answer

Expert verified

The eigenvalues areλ12=A+B±(A-B)2+4H22 which are real, and the eigenvectors areA+B±(A-B)2+4H22x+Hy=0 which are perpendicular.

Step by step solution

01

Given Information

The eigenvectors and eigenvalues for the real symmetric matrix

M=AHHB

02

Symmetric Matrix

Eigenvalues are a unique collection of scalar values associated with a set of linear equations, most commonly found in matrix equations. Characteristic roots are another name for eigenvectors. It's a non-zero vector that can only be altered by its scalar factor once linear transformations are applied.

03

Eigen Values

Mr=λrforeigenvalues,A-λHHB-λ=0A-λB-λ-H2=0λ2-(A+B)λ+AB-H2=0

λ1,2=&A+B±A+B2-4(AB-H2)2λ1,2=&A+B±A-B2+4H22

Since the matrix M is real, A , B and H are real.

The expression under the square root is positive so the eigenvalues are real.

The eigenvectors are

AHHBxy=λ1,2xy

A-B±A-B2+4H22HHA-B±A-B2+4H22xy=0A-B±A-B2+4H22x+Hy=0

This is satisfied byx=1 andA-B±A-B2+4H22 .

The eigenvectors are perpendicular by calculate their dot product

=1-A-B-A-B2+4H22H.-A-B+A-B2+4H22H=1+A-B2-A-B2+4H24H2=0

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