How do we derive the spectral projector associated with a simple eigenvalue

linear algebramarkov chainsprojection-matricesspectral-theory

Result 7.2.12 of Meyer's Matrix Analysis and Applied Linear Algebra gives the following:

If $x$ and $y^*$ are respective right and left eigenvectors of a matrix $A$ associated with a simple eigenvalue $\lambda \in \sigma(A)$, then the spectral projector associated with $\lambda$ has the form
$$G = \frac{x y^*}{y^* x}.$$

The proof that Meyer gives does not construct this form; he just verifies that it has the properties of a spectral projector. I understand that we have the spectral projector given as a normalized outer product, but I do not see from Meyer's treatment how we could derive this formula from the hypothesized eigenvectors of a simple eigenvalue.

I am interested in the result because I want to understand how the limit of the transition matrix for a primitive Markov chain gives the chain's limiting distribution as a left eigenvector. This question (Perron projection as a limit) gave me a better intuition for why $G$ works as a spectral projector of this sort, but I have not seen this projection derived.

Best Answer

Here’s how I would approach a derivation: We want the column space of $G$ to be spanned by $x$, so every column must be a multiple of $x$. This suggests the outer product of $x$ with some other vector. Similarly, we want the row space of $G$ to be spanned by $y^*$, so we try $G = \alpha xy^*$. Finally, we must have $Gx=\alpha xy^*x = x$, which gives us the normalizing factor $\alpha = (y^*x)^{-1}$. (We know that $y^*x\ne0$ because $x$ and $y^*$ share an eigenvalue.) We then have $$G^2={xy^*xy^* \over (y^*x)^2} = {xy^*\over y^*x} = G,$$ so this is indeed a projection. One must still verify that $G$ has the other desired properties, which Meyer does.

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