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I am working on imaging and I am a bit puzzled by the behaviour of this matrix:

$$A:=\left( \begin{array}{cccccc} 1 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 2 & -2 & 0 & 0 & 0 & 0 \\ -2 & 4 & -2 & 0 & 0 & 0 \\ 0 & -2 & 2 & 0 & 0 & 0 \\ \end{array} \right)$$

My matrix $A$ is a $4x4$ block matrix with an upper block $A_{11}:= \operatorname{diag}(1,0,1)$, a second $A_{12}= -id$, a lower block which is the graph Laplacian $-\Delta$ and then a block of zeros.

It is known that the lowest eigenvalue of the graph Laplacian is zero.

Now my matrix has all eigenvalues on the right hand side of the complex plane (non-negative real part) and the one with smallest real part has real part zero.

However, if I consider instead of the graph Laplacian in the $A_{21}$ block the matrix $-\Delta+id$ then this block is bounded away from zero and

$$A:=\left( \begin{array}{cccccc} 1 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 2+1 & -2 & 0 & 0 & 0 & 0 \\ -2 & 4+1 & -2 & 0 & 0 & 0 \\ 0 & -2 & 2+1 & 0 & 0 & 0 \\ \end{array} \right)$$

has only eigenvalues with strictly positive real part.

I ask: Can anybody explain the relationship between the lower left corner of my matrix $A$ having spectrum bounded away from zero and all eigenvalues of $A$ being strictly contained in the right half plane?

How does $\lambda_{\text{min}}(A_{21})$ relate to $\operatorname{min}\Re(\sigma(A))$?

EDIT: I was thinking that some Block matrix identities may be useful http://djalil.chafai.net/blog/2012/10/14/determinant-of-block-matrices/

by I do not quite get it together.

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Use that for $B=A-\lambda$ we have $$\operatorname{det}(B) = \operatorname{det}(B_{11}B_{22}-B_{12}B_{21})$$

then observe that $A$ is invertible if and only if your $A_{21}$ is (this you get by choosing $\lambda=0$. Otherwise, replace $A_{21}$ by its lowest eigenvalue and conclude the rest from the determinant formula above. This gives you the lowest eigevalue of $A$ in terms of the lowest eigenvalue of $A_{21}$-nothing mysterious here.

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