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# Open Notebook

I study computational and quantitative biology with a focus on network aging. This site is to serve as my note-book and to effectively communicate with my students and collaborators. Every now and then, a blog may be of interests to other researchers or teachers. Views in this blog are my own. All rights of research results and findings on this blog are reserved. See also http://youtube.com/c/hongqin @hongqin

## Thursday, August 8, 2019

## Wednesday, August 7, 2019

### Yuan method on adjacency matrix controllability

Yuan clearly used weighted matrix, and j->i as direction. So, column to row indicate direction?

Wikipedia, “In directed graphs, the in-degree of a vertex can be computed by summing the entries of the corresponding column, and the out-degree can be computed by summing the entries of the corresponding row.” The question now is does i->j and j->i matters? It can be tested using a star shaped network with outward and inwarding arrows. A quick exam on these show both star networks should have the same number of minimal control nodes. Well, eigen values of a matrix and its transpose are the same, see https://yutsumura.com/eigenvalues-of-a-matrix-and-its-transpose-are-the-same/.

So, i->j and j->i does not matter! This is somewhat shocking to me.

### Eigenvalues of a Matrix and its Transpose are the Same

https://yutsumura.com/eigenvalues-of-a-matrix-and-its-transpose-are-the-same/

Recall that the eigenvalues of a matrix are roots of its characteristic polynomial.

Hence if the matricesA $A$ and AT ${A}^{\mathrm{T}}$ have the same characteristic polynomial, then they have the same eigenvalues.

Hence if the matrices

So we show that the characteristic polynomial pA(t)=det(A−tI) ${p}_{A}(t)=det(A-tI)$ of A $A$ is the same as the characteristic polynomial pAT(t)=det(AT−tI) ${p}_{{A}^{\mathrm{T}}}(t)=det({A}^{\mathrm{T}}-tI)$ of the transpose AT ${A}^{\mathrm{T}}$.

We have

Therefore we obtain pAT(t)=pA(t) ${p}_{{A}^{\mathrm{T}}}(t)={p}_{A}(t)$, and we conclude that the eigenvalues of A $A$ and AT ${A}^{\mathrm{T}}$ are the same.

### Remark: Algebraic Multiplicities of Eigenvalues

Remark that since the characteristic polynomials of A $A$ and the transpose AT ${A}^{\mathrm{T}}$ are the same, it furthermore yields that the algebraic multiplicities of eigenvalues of A $A$ and AT ${A}^{\mathrm{T}}$ are the same.

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