Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Help me please to understand the formula:

Let $A$ be $n\times n$ matrix, $b$ some real number and $x$ some vector. Matrix $(A-bI)$ is $n\times n$ nonsingular matrix.

If $y_k$ are eigenvectors of the matrix $A$, then the following is true:

$$x^T(A-bI)^{-1}x=\sum_{k=1}^n \frac{(x,y_k)^2}{\lambda_k-b}$$

here $\lambda_k$ are the eigenvalues of $A$.

I do understand how we can get denominator — this is just a denominator of $(A- bI)$, because $(A-bI)^1=\det(A-bI)^{-1} \operatorname{adj}(A-bI)$.

but I don't understand how we get numerator.

Thank you

share|improve this question
    
Could you please make sure that you've written what you meant to write in that first equation? The left-hand side isn't defined (perhaps you mean to transpose one of the two appearances of $x$?) and the right-hand side doesn't have anything to do with $x$. And why is the 2 in superscript? –  Brad Nov 27 '11 at 17:38
    
Sorry, its typo the right formula is: x^T (A-bI) x= \sum_{k=1}^n \frac{(x, y_k)^2}{\lambda_k -b} thank you. –  David Nov 27 '11 at 17:52
    
If spectral for is allowed (which perhaps requires $A$ to be normal), then it becomes very easy. But I don't know about the general case. –  Tapu Nov 27 '11 at 20:33
add comment

1 Answer

up vote 0 down vote accepted

If you have the eigendecomposition $\mathbf A=\mathbf V\mathbf D\mathbf V^\top$, then $(\mathbf A-b\mathbf I)^{-1}=\mathbf V(\mathbf D-b\mathbf I)^{-1}\mathbf V^\top$, where

$$(\mathbf D-b\mathbf I)^{-1}=\mathrm{diag}\left(\frac1{d_{1,1}-b},\dots,\frac1{d_{n,n}-b}\right)$$

Thus, $\mathbf x^\top(\mathbf A-b\mathbf I)^{-1}\mathbf x=\mathbf x^\top\mathbf V(\mathbf D-b\mathbf I)^{-1}\mathbf V^\top\mathbf x=(\mathbf x^\top\mathbf V)(\mathbf D-b\mathbf I)^{-1}(\mathbf x^\top\mathbf V)^\top$. You should now be able to expand that into the sum expression you have...

share|improve this answer
    
The eigendecomposition always exists...? In my comment to the OP, I asked $A$ to be normal to ensure the spectral form. –  Tapu Nov 28 '11 at 6:56
    
Well, he certainly can't have $n$ eigenvectors just like in his summation expression if his matrix is defective... –  J. M. Nov 28 '11 at 10:41
    
Then can I assume spectral form for $A$ namely $A=\sum_k\lambda_ky_ky_k^t$...? –  Tapu Nov 28 '11 at 16:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.