# Diagonalization of Impedance Matrix

1. ## Diagonalization of Impedance Matrix

Generally an impedance matrix [Z] is a complex symmetrical matrix.
It is neither a real symmetrical matrix nor a Hermitian matrix.

However a complex impedance matrix [Z] can be diagonalized.
I don't think any impedance matrix [Z] can be diagonalized.
How can I interpret whether diagonalization is possible or not physically >

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2. ## Re: 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.

3. ## Re: Diagonalization of Impedance Matrix

Originally Posted by PlanarMetamaterials
Diagonalization is only possible iff there exists a complete set of eigenvectors.
I know such very basic thing.

Originally Posted by PlanarMetamaterials
Is [Z] a Hermitian matrix?
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 ?

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4. ## Re: Diagonalization of Impedance Matrix

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.

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5. ## Re: Diagonalization of Impedance Matrix

Originally Posted by PlanarMetamaterials
I think it may be instructive to look at diagonalization of the matrix in the first place.
Not useful at all.

6. ## Re: Diagonalization of Impedance Matrix

Code:
```Z = [2,   j,   0;
j,   2, j*2;
0, j*2,   2]```
This [Z] is neither Hermitian, Skew-Hermitian nor Unitary, yet it is Symmetric and Normal.
This [Z] can be diagonalized by Orthogonal Matrix, [P].
Code:
```[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```
Code:
```Z = [2,   j,   0;
j,   2, j*2;
0, j*2,   2.1]```
This [Z] is also neither Hermitian, Skew-Hermitian nor Unitary, yet it is Symmetric.
However [Z] is not Normal.

This [Z] still can be diagonalized by [P]
However [P] is neither Unitary nor Orthogonal Matrix.
Code:
```[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```
How can I interpret ?

7. ## Re: Diagonalization of Impedance Matrix

[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.

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8. ## Re: Diagonalization of Impedance Matrix

Just for example.

For series RC circuit, [Z] can not be diagonalized at omega=2/(R*C).

9. ## Re: Diagonalization of Impedance Matrix

Originally Posted by LMA
For series RC circuit, [Z] can not be diagonalized at omega=2/(R*C).
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.

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