# Orthogonal in the B Norm?

If you have two generalized eigenvectors $\varphi_1 , \varphi_2$ (with different eigenvalues) of a matrix A, then they will be orthogonal in the B norm.

In this context, I do not understand what is meant by the "B norm" where B is a matrix of the same dimensions as A. What does it mean to be orthogonal in another matrices' norm?

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1
Could you provide a reference for this? It is probably easier to figure it out in context. – Calle May 29 '11 at 18:27

## 1 Answer

You have $A \varphi_i = \lambda_i B \varphi_i$. I'm assuming $A$ and $B$ are symmetric, with $B$ positive definite. Then $\lambda_1 \varphi_1^T B \varphi_2 = \varphi_1^T A \varphi_2 = \lambda_2 \varphi_1^T B \varphi_2$ with $\lambda_1 \ne \lambda_2$, so $\varphi_1^T B \varphi_2 = 0$. This says that $\varphi_1$ and $\varphi_2$ are orthogonal in the inner product $(u,v) = u^T B v$ corresponding to the matrix $B$, which might be abbreviated as "in the $B$ norm".

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Thanks for the information! I'm trying to parse your response because I'm a novice in Linear Algebra. – sam May 29 '11 at 18:39
When you say that, "$\varphi_1$ and $\varphi_2$ are orthogonal in the inner product $(u,v)=u^TBv$", are $u,v$ just $\varphi_1 , \varphi_2$? – sam May 29 '11 at 18:53
In the definition of the inner product, $u$ and $v$ are any vectors in the vector space.  You are using this definition with $u = \varphi_1$, $v = \varphi_2$. – Robert Israel May 30 '11 at 2:23
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