EMLA Q.46

  • Option : A
  • Explanation : Let A be a real symmetric matrix, therefore

    AT=A

    Let α1 and α2 be different eigen values of matrix A, and X1 and X2 be the corresponding vectors, then

    AX1= α1X1 and AX2 = α2X2

    Taking transpose of the second equation 

    (AX2)T=  (α2X2)

    X2TAT= α2.X2T 2X2T

    But AT-A

    Linear Algebra

    Post multiply by X1, we get

    XT2AX1 = a2 X2T X1

    But AX1 = a1X1

      XT2 a1X1 = a2 X2T X1

     (a1 - a2) X2TX1 = 0

    Since a1  a2, a1 - a2  0

     X2TX1 = 0 i.e. X2 and X1 are orthogonal.

Cancel reply

Your email address will not be published. Required fields are marked *


Cancel reply

Your email address will not be published. Required fields are marked *