Diagonalization of Impedance Matrix

Diagonalization is only possible iff there exists a complete set of eigenvectors.

Is [Z] a Hermitian matrix?

See Horn and Johnson's Matrix Analysis.

I know such very basic thing.

No. [Z] is not a Hermitian matrix.

Here [Z]^T means transpose of [Z], [Z]⁺ means conjugate transpose of [Z].

For lossless reciprocal circuit, [Z] is a pure imaginary symmetrical matrix which is a skew Hermitian Matrix at same time.
[Z]^T==[Z], [Z]⁺==-[Z]
[Z]^T*[Z]==[Z]*[Z]^T, [Z]⁺*[Z]==[Z]*[Z]⁺
In this case, eigenvectors are real, eigenvalues are pure imaginary.
We can always diagonalize [Z] by Orthogonal Matrix. It is not by Unitary Matrix.

On the other hand, for reciprocal circuit with loss, [Z] is neither a real symmetrical matrix nor Hermitian Matrix. It is complex symmetrical matrix.
[Z]^T==[Z], [Z]⁺!=[Z]
[Z]^T*[Z]==[Z]*[Z]^T, [Z]⁺*[Z]!=[Z]*[Z]⁺
However we can still diagonalize [Z] by Orthogonal Matrx. It is not by Unitary Matrix.

Of course, if eigenvectors are not linearly independent, diagonalization is impossible.

How can I interpret whether diagonalization is possible or not physically ?

I think it may be instructive to look at diagonalization of the matrix in the first place.

If we have such a diagonalization, then: [Z]I = Iλ, with λ being the eigenvalues and I being the currents. This means that, since the impedance matrix definition is V = [Z]I, that:

V = Iλ.

I.e., there exists a certain set of currents for which all of the corresponding voltages are related by the same value (λ, or if you prefer, a modal impedance Zm).

As to what this means physically.... good question. A series of impedances connected to all ports might do it; but I can't think what purpose it might serve.

Not useful at all.

Z = [2,   j,   0;
     j,   2, j*2;
     0, j*2,   2]

[P, D] = eig(Z)

P =

   0.3162 + 0.0000i   0.8944            -0.3162          
   0.7071             0.0000 - 0.0000i   0.7071          
   0.6325 - 0.0000i  -0.4472 + 0.0000i  -0.6325 - 0.0000i


D =

   2.0000 + 2.2361i        0                  0          
        0             2.0000 - 0.0000i        0          
        0                  0             2.0000 - 2.2361i

Z = [2,   j,   0;
     j,   2, j*2;
     0, j*2,   2.1]

[P, D] = eig(Z)

P =

   0.3162 + 0.0057i   0.8942            -0.3162 + 0.0057i
   0.7071             0.0000 - 0.0179i   0.7071          
   0.6322 - 0.0170i  -0.4473 - 0.0000i  -0.6322 - 0.0170i


D =

   2.0400 + 2.2354i        0                  0          
        0             2.0200 + 0.0000i        0          
        0                  0             2.0400 - 2.2354i

[P, D] = eig(Z)


(Case-1)
Z = [1, j; j, 1]
Z : Symmetric=1
Z : Hermitian=0
Z : Skew-Hermitian=0
Z : Orthogonal=0
Z : Unitary=0
Z : Normal=1

P = [0.7071,0.7071; 0.7071,-0.7071]
P : Orthogonal=1
P : Unitary=1
rank(P)=2
det(P)=-1+j*-5.88785e-017
cond(P)=1

D = [1.0000+1.0000*j,0; 0,1.0000-1.0000*j]


(Case-2)
Z = [1, j; j, 2]
Z : Symmetric=1
Z : Hermitian=0
Z : Skew-Hermitian=0
Z : Orthogonal=0
Z : Unitary=0
Z : Normal=0

P = [0.7071,-0.6124+0.3536*j; 0.6124-0.3536*j,0.7071]
P : Orthogonal=0
P : Unitary=0
rank(P)=2
det(P)=0.75+j*-0.433013
cond(P)=1.73205

D = [1.5000+0.8660*j,0; 0,1.5000-0.8660*j]


(Case-3)
Z = [1, j; j, 3]
Z : Symmetric=1
Z : Hermitian=0
Z : Skew-Hermitian=0
Z : Orthogonal=0
Z : Unitary=0
Z : Normal=0

P = [0.7071,0.7071*j; -0.7071*j,0.7071]
P : Orthogonal=0
P : Unitary=0
rank(P)=1
det(P)=2.10734e-008+j*0
cond(P)=9.49063e+007

D = [2.0000,0; 0,2.0000]

Eigenvalues are degenerated and Eigenvectors are not linearly independent in Case-3.

Just for example.

Thanks for very good example.

Why is a diagonalization impossible in this frequency ?
What does this frequency mean?
This frequency is twice of 3-dB cutoff frequency.