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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 (112k points) | 3.3k views
+2
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).
0

It is a question from basis for each eigenspace .

0

@ maam

what is meaning of this in above link (Ax,y) , (x,Ay) 

0

It is property of eigen vector

if Ax = λ x

then $x^{T} A x = \lambda x^{T} x = \lambda (1) = \lambda$

4 Answers

+41 votes
Best answer

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 (12.3k points)
selected ago by
+2
thnx for the explanation.
+1
This should be accepted answer.
+1
what an explanation, man
+2

@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.

+34 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
answered by Loyal (7.7k points)
+3
Can someone please explain it in easy way  .. I am not getting  the explanation  of stackexchange .

Thank you
+1 vote
answered by Loyal (8.2k points)
0
How the normalisation has taken place?
–5 votes
1

 

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

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