# Thread: Hermitian scalar product relationship

1. ## Hermitian scalar product relationship

Hi all i could use a little help. Thanks in advance.

Prove that if A is Hermitian than (Ax, y) = (x, Ay), where x and y are any vectors.

what ive tried:

(Ax, y) = summation(A.x_i . conj(y_i); i = 1...n)

(x, Ay) = summation(x_i . conj(A . y_i); i =1...n) but A = A* so
(x, Ay) = summation(x_i . conj(A*, y_i); i =1...n) and A* = conj(trans(A))
note*[trans = transpose]
(x, Ay) = summation(x_i . conj(conj(trans(A)), y_i); i =1...n) which can be simplified to
(x, Ay) = summation(x_i . trans(A) . conj(y_i); i =1....n) using distributive property of matrix multiplication
(x, Ay) = summation(trans(A) . x_i . conj(y_i); i =1....n)

I must be missing something because i cannot guarantee A = trans(A) or A = conj(A) i think.

Thanks.

2. Originally Posted by OkashiiKen
Hi all i could use a little help. Thanks in advance.

Prove that if A is Hermitian than (Ax, y) = (x, Ay), where x and y are any vectors.

what ive tried:

(Ax, y) = summation(A.x_i . conj(y_i); i = 1...n)

(x, Ay) = summation(x_i . conj(A . y_i); i =1...n) but A = A* so
(x, Ay) = summation(x_i . conj(A*, y_i); i =1...n) and A* = conj(trans(A))
note*[trans = transpose]
(x, Ay) = summation(x_i . conj(conj(trans(A)), y_i); i =1...n) which can be simplified to
(x, Ay) = summation(x_i . trans(A) . conj(y_i); i =1....n) using distributive property of matrix multiplication
(x, Ay) = summation(trans(A) . x_i . conj(y_i); i =1....n)

I must be missing something because i cannot guarantee A = trans(A) or A = conj(A) i think.

Thanks.
A matrix (operator) A is Hermitian iff $A^{\dagger} = A$.

So given two column vectors x, y:
$(Ax, y) \equiv x^{\dagger} \cdot A^{\dagger}y = \sum_j x_j^* (A^{\dagger}y)_j$

But $A^{\dagger} = A$ so:
$(Ax, y) \equiv x^{\dagger} \cdot A^{\dagger}y = \sum_j x_j^* (A^{\dagger}y)_j$ $= \sum_j x_j^* (Ay)_j = \sum_{ij} x_j^* A_{ji}y_i$

Similarly:
$(x, Ay) = x^{\dagger} \cdot Ay = \sum_{j}x_j^*(Ay)_j = \sum_{ij} x_j^* A_{ji}y_i$

Since these two expressions are equal, (Ax, y) = (x, Ay).

-Dan

3. Originally Posted by OkashiiKen
Hi all i could use a little help. Thanks in advance.

Prove that if A is Hermitian than (Ax, y) = (x, Ay), where x and y are any vectors.

what ive tried:

(Ax, y) = summation(A.x_i . conj(y_i); i = 1...n)

(x, Ay) = summation(x_i . conj(A . y_i); i =1...n) but A = A* so
(x, Ay) = summation(x_i . conj(A*, y_i); i =1...n) and A* = conj(trans(A))
note*[trans = transpose]
(x, Ay) = summation(x_i . conj(conj(trans(A)), y_i); i =1...n) which can be simplified to
(x, Ay) = summation(x_i . trans(A) . conj(y_i); i =1....n) using distributive property of matrix multiplication
(x, Ay) = summation(trans(A) . x_i . conj(y_i); i =1....n)

I must be missing something because i cannot guarantee A = trans(A) or A = conj(A) i think.

Thanks.
$(Ax,b)=\sum_i \left(\sum_j (A_{i,j}x_j) \right) \overline{y_i}$
$= \sum_i \left(\sum_j (A_{i,j}x_j) \overline{y_i}\right)$
$= \sum_i \left(\sum_j x_j \overline{\overline{A_{i,j}}y_i}\right)$

Now change the order of summation:

$(Ax,b)= \sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= (x,\overline{A}^{\,t} y) =(x,A^H y)$

and since $A$ is Hermitian $A=A^H$ so:

$(Ax,b) =(x,A y)$.

RonL

4. Originally Posted by topsquark
A matrix (operator) A is Hermitian iff $A^{\dagger} = A$.

So given two column vectors x, y:
$(Ax, y) \equiv x^{\dagger} \cdot A^{\dagger}y = \sum_j x_j^* (A^{\dagger}y)_j$

But $A^{\dagger} = A$ so:
$(Ax, y) \equiv x^{\dagger} \cdot A^{\dagger}y = \sum_j x_j^* (A^{\dagger}y)_j$ $= \sum_j x_j^* (Ay)_j = \sum_{ij} x_j^* A_{ji}y_i$

Similarly:
$(x, Ay) = x^{\dagger} \cdot Ay = \sum_{j}x_j^*(Ay)_j = \sum_{ij} x_j^* A_{ji}y_i$

Since these two expressions are equal, (Ax, y) = (x, Ay).

-Dan
Thanks, but i am having a little difficulty following.... :/

i dont understand this operator:
$A^{\dagger} = A$
I had understood a hermitian matrix to be one such that
$A* = A$ and $a_{ij} = \overline{a_{ji}}$

Your first step made some leaps i cant quite seem to follow especially this one:
$(x, Ay) = x^{\dagger} \cdot Ay = \sum_{j}x_j^*(Ay)_j = \sum_{ij} x_j^* A_{ji}y_i$
I can mostly follow after this step, but I dont understand this equation. Could you elaborate this step for me? Thank you.

Originally Posted by CaptainBlack
$(Ax,b)=\sum_i \left(\sum_j (A_{i,j}x_j) \right) \overline{y_i}$
$= \sum_i \left(\sum_j (A_{i,j}x_j) \overline{y_i}\right)$
$= \sum_i \left(\sum_j x_j \overline{\overline{A_{i,j}}y_i}\right)$

Now change the order of summation:

$(Ax,b)= \sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= (x,\overline{A}^{\,t} y) =(x,A^H y)$

and since $A$ is Hermitian $A=A^H$ so:

$(Ax,b) =(x,A y)$.

RonL
This solution is much more clear to me. Although i am still confused at one step.
$(Ax,b)= \sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= (x,\overline{A}^{\,t} y) =(x,A^H y)$
I cant follow this:
$\sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= (x,\overline{A}^{\,t} y)$
When i work it out:
$\sum_j \left(\sum_i x_j \overline{\overline{A_{i,j}}y_i}\right)= \sum_j \left(\sum_i x_j A_{i,j}\overline y_i\right)=(x,\overline A y)$
Am I missing the transpose because we are summing across the i direction instead of the j direction? I have a problem when working from (x, Ay)

I keep showing that
$(Ax, y) = (x, A^T y)$

Thanks again.

5. Originally Posted by OkashiiKen
Thanks, but i am having a little difficulty following.... :/

i dont understand this operator:
$A^{\dagger} = A$
I had understood a hermitian matrix to be one such that
$A* = A$ and $a_{ij} = \overline{a_{ji}}$

Your first step made some leaps i cant quite seem to follow especially this one:
$(x, Ay) = x^{\dagger} \cdot Ay = \sum_{j}x_j^*(Ay)_j = \sum_{ij} x_j^* A_{ji}y_i$
I can mostly follow after this step, but I dont understand this equation. Could you elaborate this step for me? Thank you.
I could elaborate, but I suspect you will be more comfortable following CaptainBlack's solution. We are essentially saying the same thing, just in a different notation. I am not as familiar with the notation you and CaptainBlack are using, so I'll leave it to him.

For the sake of clarity I will mention that the $\dagger$ operator is defined (virtually always, at least in Physics) as the Hermitian conjugate, or transpose conjugate, operator: Acting on a square matrix A, $A_{ij}^{\dagger} = A_{ji}^*$. (Where, to be clear, * is the complex conjugate operator.) So the Hermitian scalar product acting on two vectors x (a row vector) and y (a column vector) is $(x, y) = x^{\dagger} \cdot y$, where $\cdot$ is the usual (Euclidean) scalar product.

-Dan

6. Thanks Topsquark.

I feeling pretty slow today....forgot that

$x^T y = \sum_i x_i y_i$

Thank you CaptainBlack i can follow your solution now! Both of you were a big help. Now i can move on to my next problem...proving all eigenvalues of a hermitian matrix A are real. Thanks for all your help.