# Image Matrix Vector Representation for the Degradation Model

I am trying to understand the the degradation model equation but I have doubt that how come y^t.x.h will be equal to x^t.h^t.y . Aren't they transpose of each other. • Could you please review my answer and mark it? – Royi Dec 16 '20 at 5:21

## 2 Answers

Indeed since both expressions are scalars then they are equal to each other since the transpose of a scalar is the same scalar.

See in MATLAB as an example (Calculating $${x}^{T} H y$$ and $${y}^{T} {H}^{T} x$$:

>> vX = randn(10, 1);
>> vY = randn(10, 1);
>> mH = randn(10, 10);
>> vX.' * mH.' * vY

ans = -0.8618

>> vY.' * mH * vX

ans = -0.8618



As you can see, indeed both expressions are scalars.
As expected $${x}^{T} H y = {\left( {x}^{T} H y \right)}^{T} = {y}^{T} {H}^{T} x$$.

Here the quantity $$y^T H x$$ is a real scalar (corresponding to an energy). And scalars can also be considered as matrices of dimension $$1 \times 1$$. Such matrices are equal to their transpose.

When you have a product of matrices that ends up in a scalar, one often derives results using the associativity of the matrix products, and the property of transposition: $$(AB)^T=B^TA^T$$. Steps by steps, you have:

$$(y^T H x)^T = (y^T (H x))^T = (H x)^T (y^T)^T= x^TH^T(y^T)^T$$

thus

$$(y^T H x)^T = x^TH^Ty\,.$$

Warning: when you have complex quantities, you have to play with complex conjugate has well.

• Thanks @Laurent Duval – SOMA REDDY Dec 17 '20 at 8:19
• I have added a couple of intermediate computations, please tell me whether is it useful – Laurent Duval Dec 17 '20 at 8:53