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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.

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    $\begingroup$ Could you please review my answer and mark it? $\endgroup$ – Royi Dec 16 '20 at 5:21
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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 $.

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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.

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

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