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I'm currently examining irreducible diagonally dominant matrices say A, i.e. $\exists i$ $|a_{ii}| > \sum _{i \not = j} a_{ij}$ and the corresponding graph G to the adjacency matrix A is strongly connected (to have an irreducible matrix).

An example is the following matrix: $\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$ with the eigenvalues $\lambda_1=1$, $\lambda_2=3$.

Does the smallest eigenvalue always decrease when a diagonal element decreases by 1 in this class of systems? For example $\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$ has the eigenvalues $\lambda_1 \approx 0.38$ and $\lambda_2 \approx 2.62$. The Gershgorin circle theorem doesn't say anything because the circles overlap, so I'm curious if it's possible to show something for this problem. Clearly the sum of all eigenvalues equal the trace of A, but can this also be related to the smallest eigenvalue in some way?

Greetings, Vincent

whpowell96
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zinsinho
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  • Are your matrices always symmetric in addition to these other properties? If so, there are more precise estimates for perturbations of eigenvalues. – whpowell96 May 22 '24 at 17:14
  • Let $A$ be your original matrix, and $A' = A-\mathbf e_k\mathbf e_k^T$ with eigenvalues $\lambda_i$ and $\lambda_i'$ respectively (always sorted in descending order) then $A\succeq A'\implies \lambda_i \geq \lambda_i'$. The fact that you are interested in the smallest eigenvalue only means you should be able to directly prove the result with a simple quadratic form argument, i.e. choose appropriate $\mathbf x$ so $\lambda_n = \mathbf x^T A \mathbf x\geq \mathbf x^T A' \mathbf x\geq \lambda_n'$ – user8675309 May 22 '24 at 18:51
  • @whpowell96 the matrices I want to consider do not have to be symmetric. However, I think it's good for me to understand what applies to symmetric matrices. – zinsinho May 23 '24 at 05:56

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Since you are discussing adjacency matrices, I will assume that all of these matrices are symmetric. Let $A$ be a $n\times n$ real and symmetric. It then has $n$ real eigenvalues which we order like so: $\lambda_n \leq \lambda_{n-1} \leq\dots\leq \lambda_1$. The Courant-Fischer Minimax Theorem then states the following:

$$ \lambda_k(A) = \max_{\dim(S)=k}~\min_{0\neq y\in S}~ \frac{y^TAy}{y^Ty}, $$ or, equivalently, $$ \lambda_k(A) = \max_{\dim(S)=k}~\min_{y\in S, \ \|y\|=1}~ y^TAy. $$

Now consider the case for the smallest eigenvalue, which is when $k=n$. In this case, all $k$-dimensional subspaces are just equal to $\mathbb{R}^n$, so the maximum is trivially satisfied and we obtain a nicer result:

$$ \lambda_n(A) = \min_{y\in \mathbb{R}^n, \ \|y\|=1}~ y^TAy. $$

Now consider a negative semidefinite rank-1 perturbation of $A$, $B = A + \gamma uu^T$, where $\gamma<0$ and $u\neq 0$. In the case where $\gamma = -1$ and $u = e_k$, this corresponds to subtracting $1$ from the $k$-th diagonal element. From the previous characterization, we have $$ \lambda_n(B) = \min_{y\in \mathbb{R}^n, \ \|y\|=1}~ y^TBy = \min_{y\in \mathbb{R}^n, \ \|y\|=1}~ y^TAy + \gamma y^Tuu^Ty = \min_{y\in \mathbb{R}^n, \ \|y\|=1}~ y^TAy + \gamma (y^Tu)^2 \leq \min_{y\in \mathbb{R}^n, \ \|y\|=1}~ y^TAy = \lambda_n(A). $$ This shows that the smallest eigenvalue cannot increase and that is the best you can do in general. For example, if $u$ is orthogonal to the eigenspace corresponding to $\lambda_n(A)$ and $|\gamma|<|\lambda_n(A) - \lambda_{n-1}(A)|$ then the smallest eigenvalue can remain unchanged. For example, $A = \begin{pmatrix}3&0\\0&1\end{pmatrix}$ where $1$ is subtracted from the first diagonal element.

whpowell96
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  • Thank you! Do you have any literature recommendations to go in more detail about the used theorem? Is there something relatable for non-symmetric matrices ? – zinsinho May 23 '24 at 06:11
  • The only thing I brag about is that the example used is not an irreducible matrix. At the moment I have not found any example of an irreducible matrix A where the smallest eigenvalue remains unchanged when a diagonal element is decreasing. – zinsinho May 23 '24 at 06:17
  • I don’t know much about irreducible matrices but my gut tells me that $e_k$ can’t be orthogonal to their eigenspaces. As for a reference, check out Golub and Van Loan – whpowell96 May 23 '24 at 11:08
  • Is my understanding in the stated (symmetric) case correct, that the smallest eigenvalue can only remain unchanged if u is orthogonal to y and y beeing the corresponding eigenvector to the smallest eigenvalue ? – zinsinho May 23 '24 at 12:07
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    The following definition holds for a irreducible matrix $A$ : "$A$ does not have non-trivial invariant coordinate subspaces. Here a non-trivial coordinate subspace means a linear subspace spanned by any proper subset of standard basis vectors of $F^n$. More explicitly, for any linear subspace spanned by standard basis vectors $e_{i1},...,e_{ik}$ , $0<k<n$ its image under the action of $A$ is not contained in the same subspace". I guess from this I can directly conclude that $e_k$ can not be orthogonal to the eigenspace of an irreducible matrix, so the inequality turns into a strict inequality. – zinsinho May 23 '24 at 13:03