geometric multiplicity and algebraic multiplicity are equal for symmetric matrice

𝜆$\lambda$ be an eigen value of a symmetric matrix 𝐴$A$ then how to show that the geometric multiplicity and algebraic multiplicity are equal?

in general, geometric multiplicity <= algebraic multipliciy

https://youtu.be/Xcln3xG8QGQ

Matrix Eigen value

spectral theorem

https://youtu.be/KCANLl8z6PI

visual explanation
https://youtu.be/PFDu9oVAE-g

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

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

pAT(t)=det(AT−tI)=det(AT−tIT)=det((A−tI)T)=det(A−tI)=pA(t).since IT=Isince det(BT)=det(B) for any square matrix B$$\begin{array}{rl}& p_{A^{\mathrm{T}}}(t)=det(A^{\mathrm{T}}-tI)\\ & =det(A^{\mathrm{T}}-t{I}^{\mathrm{T}})& & \text{since }{I}^{\mathrm{T}}=I\\ & =det\left((A-tI{)}^{\mathrm{T}}\right)\\ & =det(A-tI)& & \text{since }det(B^{\mathrm{T}})=det(B)\text{ for any square matrix }B\\ & =p_A(t).\end{array}$$

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.