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+20 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 ___________
asked in Linear Algebra by Veteran (99.8k points) | 2.9k views
Guys! I don't have precise explaination to this answer but I do have an imprecise one. Please discuss and help me out with the explaination if you can. My explaination to this goes as follows:

Eigen vector is a vector quantity, while eigen value is a scalar quantity. We perform dot product between two vectors and [A - ^ * I] is a vector( where ^ (lambda) is eigen value). So in my view the question is asking for a multiplication between X(eigen vector) and [A - ^ * I].

Further by looking at the answer key from GATE 14 I found it says 0, dot product of any two vector can be zero if angle between the two vectors is 90(Pi/4).

I am in a hanging garden for this one, if any one have better explaination please put it through.


Answer: 0 ( I just have a Hazzy idea).

It is a question from basis for each eigenspace .

4 Answers

+33 votes
Best answer
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:
answered by Loyal (7.6k points)
selected by
Can someone please explain it in easy way  .. I am not getting  the explanation  of stackexchange .

Thank you
+30 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.

answered by Boss (11.9k points)
thnx for the explanation.
This should be accepted answer.
what an explanation, man
vTu      is nothing but the dot product of the eigen vectors u and v

can someone please explain this line?

@janeb abhishek

vΤu is a dot product because, v and u are column vectors of n-by-1. So vΤ is row vector of 1-by-n. So multiplying these two will give 1-by-1 value, which is actually the dot product.

+1 vote
answered by Loyal (7.8k points)
How the normalisation has taken place?
–5 votes


eigon vectors are orthogonal .
answered by Junior (619 points)

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