EMLA Q.46

0. The eigen vectors of a real symmetric matrix corresponding to different eigen values are

Answer
Comment
Array

Option : A
Explanation : Let A be a real symmetric matrix, therefore
A^T=A
Let α₁and α₂ be different eigen values of matrix A, and X₁and X₂be the corresponding vectors, then
AX₁= α₁X₁and AX₂ = α₂X₂
Taking transpose of the second equation
(AX₂)^T= (α₂X₂)
X_2TA^T= α₂.X_{2T 2}X_2T
But A^T-A
Post multiply by X₁, we get
X^T₂AX₁ = a₂X_2TX₁
But AX₁ = a₁X₁
$\therefore$ X^T₂a₁X₁ = a₂X_2TX₁
$\Rightarrow$ (a₁- a₂) X_2TX₁ = 0
Since a₁ $\neq$ a2, a₁- a₂ $\neq$ 0
$\therefore$ X_2TX₁ = 0 i.e. X₂and X₁are orthogonal.

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