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60 votes
60 votes
The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a $4-by-4$ symmetric positive definite matrix is ___________

4 Answers

Best answer
146 votes
146 votes

Let $\lambda_{1}$ and $\lambda_{2}$ be two distinct eigenvalues of matrix $A$ and $u$ and $v$ be their corresponding eigenvectors respectively.

We know that an eigenvector $X$ corresponding to an eigenvalue $\lambda$ of matrix $A$ satisfies 

$AX = \lambda X$

$\therefore Au = \lambda _{1}u \quad \to (1)$ and $Av = \lambda _{2}v \quad\to (2)$

On pre-multiplying eqn $(1)$ with $v^{T}$, we get

$v^{T}Au = v^{T}\lambda _{1}u$

$\left ( v^{T}A\right )u = v^{T}\lambda _{1}u$

$\left ( A^{T}v\right )^{T}u = v^{T}\lambda _{1}u$

 $\left ( Av\right )^{T}u = v^{T}\lambda _{1}u$  $($since $A$ is a symmetric matrix, we can write $A^{T} = A)$

But $Av = \lambda _{2}v$ ... from $(2)$

$\left ( \lambda _{2}v\right )^{T}u = v^{T}\lambda _{1}u$

$\lambda _{2}v^{T}u = v^{T}\lambda _{1}u$

$\lambda _{2}v^{T}u = \lambda _{1}v^{T}u$ $($as $\lambda _{1}$ is a constant, we can write $v^{T}\lambda _{1} = \lambda _{1}v^{T})$

$\lambda _{2}v^{T}u - \lambda _{1}v^{T}u = 0$

$\left ( \lambda _{2} - \lambda _{1} \right )v^{T}u = 0$

$\therefore$ Either $\left ( \lambda _{2} - \lambda _{1} \right ) = 0$ or $v^{T}u = 0$

But since $\lambda _{2} \neq \lambda _{1}$,  $v^{T}u$ must be $0$

$v^{T}u$ is nothing but the dot product of the eigenvectors $u$ and $v$

Hence, we can conclude that the eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are orthogonal.

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56 votes
56 votes
Answer to this question is ZERO.

This is because eigen vectors corresponding to DIFFERENT eigen values of a REAL symmetric matrix are ORTHOGONAL to each other.

However, same eigen values they may not be.

And Dot -product of orthogonal vectors(perpendicular vectors ) is 0 (ZERO)

For more info see the link: http://math.stackexchange.com/questions/82467/eigenvectors-of-real-symmetric-matrices-are-orthogonal
–4 votes
–4 votes
1

 

eigon vectors are orthogonal .
Answer:

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