0
$\begingroup$

Help! It's been almost 20 years since I did much linear algebra and I'm struggling to follow a derivation. In the following expression $$\mathrm{y}' = A(A + \beta I)^{-1}\mathbf{y}$$ the real symmetric matrix $A$ has the eigenvector decomposition $$A = \sum^n_{i=1} \lambda_i\mathbf{u}_i\mathbf{u}_i^T$$ Vector $\mathbf{y}$ can be represented as a linear combination of $A$'s eigenvectors: $$\mathbf{y} = \sum^n_{i=1}\alpha_i \mathbf{u}_i$$ where $\alpha_i = \mathbf{u}_i^T \mathbf{y}$. This much I'm happy with, but what I don't understand is how one then gets to $$\mathbf{y}' = \sum^n_{i=1} \frac{\lambda_i\alpha_i}{\lambda_i + \beta}\mathbf{u}_i$$ Can somebody offer a clear explanation for this step? Is there a way to show that $$(A + \beta I)^{-1} = \sum^n_{i=1}\frac{1}{\lambda_i + \beta}\mathbf{u}_i\mathbf{u}_i^T$$ or is there something else going on?

$\endgroup$
  • $\begingroup$ I expect the somewhere along the way the derivation exploits the mutual orthogonality of the eigenvectors. $\endgroup$ – amd Sep 27 '17 at 23:39
  • $\begingroup$ Looks like my question is related to math.stackexchange.com/questions/237871/… and requires knowledge that $A+xI$ has same eigenvectors as $A$ with eigenvalues $\lambda+x$, and that $A^{-1}$ same eigenvectors as $A$ with eigenvalues $1/\lambda$ $\endgroup$ – beldaz Sep 28 '17 at 1:38
0
$\begingroup$

I ended up finding out the answers for myself, so recording here for posterity.

Firstly, recall that one definition for an eigenvector $v$ with eigenvalue $\lambda$ of a matrix $A$ is that $$(A-\lambda I)v = 0$$ The eigenvectors provide an orthonormal set of vectors, so since $$\mathbf{y} = \sum^n_{i=1}\alpha_i \mathbf{u}_i$$

$$(A+\beta I)\mathbf{y} = A\mathbf{y}+\beta I\mathbf{y} = \sum_{i=1}^n\sum_{j=1}^n \alpha_j \lambda_i \mathbf{u}_i \mathbf{u}_i^T \mathbf{u}_j + \beta\sum_{i=1}^n\alpha_i \mathbf{u}_i$$

and since because of the orthonormal property $\mathbf{u}_i^T \mathbf{u}_j = 1$ for $i=j$ and 0 otherwise, this reduces to

$$(A+\beta I)\mathbf{y} = \sum_{i=1}^n \alpha_i(\lambda_i + \beta)\mathbf{u}_i$$

so $(A+\beta I)$ is equivalent to a matrix $A'$ with the same eigenvectors as in $A$, but with eigenvalues $\lambda_i' = \lambda_i + \beta$.

Secondly, as shown in this MSE question the eigenvectors of a matrix $M$ are the same as those for its inverse $M^{-1}$, with reciprocal corresponding eigenvalues. So $$M^{-1} = \sum_{i=1}^n \frac{1}{\lambda_i} \mathbf{u}_i \mathbf{u}_i^T$$ where $\mathbf{u}_i$ and $\lambda_i$ are the $i$th eigenvector and eigenvalue, respectively, of $M$. So using the above composite matrix $A'=A+\beta I$ for $M$, we have $$(A+\beta I)^{-1} = \sum_{i=1}^n \frac{1}{\lambda_i + \beta} \mathbf{u}_i \mathbf{u}_i^T$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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