# What is the a matrix decomposition to swap out an eigenvalue?

I have the following problem: let $$A$$ be an $$m \times m$$ matrix that is not symmetric and real. Then, I am interested in its eigenspaces that correspond to the largest eigenvalue, $$\lambda_{\text{max}}$$. Without affecting the column span of the eigenspace associated with the largest root, say $$\text{sp}\,r_{\text{max}}$$, what can I do to $$A$$ to effect the following change: 1. remove $$\lambda_{\text{max}}$$, 2. insert $$\mu$$ in its stead, 3. obtain the relation $$A\, r_{\text{max}} = \mu r_{\text{max}}$$ so that the eigenvalue associated with $$\text{sp}\,r_{\text{max}}$$ is now $$\mu$$. I was thinking of possibly using the real Schur decomposition, thanks to its numerical stability. Any help is much appreciated!

Follow-up: suppose that the eigenvalue is potentially complex but the matrix is real. Then, as suggested, calculating $$A + I \left(\mu -\lambda_{\text{max}} \right)$$ would cause a problem because $$\lambda_{\text{max}}$$ comes in a conjugate pair. We (re-)define the notion of 'max' to be maximal by modulus, so $$\lambda_{\text{max}}$$ is such that $$||\lambda_{\text{max}}||$$ is maximal. (This arises from the fact that I am estimating $$A$$ using least squares. A few references, avenues I have considered is applying the method of Gershgorin discs or the Bauer-Fike theorem just to bound $$||\lambda_{\text{max}}||$$ using the (real) entries of $$A$$ but have not made much progress.

Moreover, it would be interesting to know if one could use the derivative of $$A$$ with respect to its eigenvalues to 'slide it' towards a matrix that would have the desired root, very much one uses the first derivative of a function to attain a linear approximation to.

## 1 Answer

The "obvious" thing to do is to change each $$\lambda_{max}$$ on the diagonal of the Schur decomposition to $$\mu$$, but that may change the eigenspace. For example, if $$A = \pmatrix{1 & 1\cr 0 & 2\cr}$$ (already upper triangular) the eigenspace for $$2$$ is spanned by $$\pmatrix{1\cr 1\cr}$$, but if you change the $$2$$ to $$\mu$$ the eigenspace for $$\mu$$ becomes $$\pmatrix{1/(\mu-1)\cr 1\cr}$$.

Instead, why not just add $$(\mu - \lambda_{max}) I$$ to the matrix? The eigenspaces stay the same, just that all eigenvalues are shifted by $$\mu - \lambda_{max}$$.

EDIT: Another possibility is to use the functional calculus: for any polynomial $$p(X)$$, $$p(A)$$ is a matrix whose eigenvalues are $$p(\lambda)$$ for eigenvalues $$\lambda$$ of $$A$$. If $$v$$ is an eigenvector of $$A$$ for $$\lambda$$, then $$p(A) v = p(\lambda) v$$. Moreover, if $$p$$ has real coefficients, $$p(A)$$ will still be real. Choose a polynomial such that $$p(\lambda_{max}) = \mu$$.

• Thank you so much for your help @RobertIsrael. Just a quick follow-up, this requires the eigenvalues to be real, and not complex, right? I tried working with potentially complex eigenvalues (which I understand weren't part of my original question), but ran into trouble then. – Hirek Aug 14 at 16:02
• Why would it require that? What kind of trouble? – Robert Israel Aug 14 at 17:56
• Hi @RobertIsrael I have added a little follow-up to my question. Anyway, thank you so much for your kind attention! – Hirek Aug 14 at 18:08
• Thanks so much, but just to make sure I understand, did you mean to write if $v$ is an eigenvector of $A$? – Hirek Aug 15 at 16:17
• If $Av = \lambda v$ then by induction $A^n v = \lambda^n v$ for all nonnegative integers $n$ (including $n=0$ where $A^0 = I$), and so if $p(x) = \sum_{k=0}^n c_k x^k$ is a polynomial, $p(A) v = \sum_{k=0}^n c_k A^k v = \sum_{k=0}^n c_k \lambda^k v = p(\lambda) v$. – Robert Israel Aug 15 at 17:42