Transpose of the product of matrices problem

[EdMel]

Hi,

The following equations are from linear regression model notes but there is an aspect of the matrix algebra I do not get.

I have, \mathbf{y} and \tilde{\beta} are a mx1 vectors, and \mathbf{X} is a mxn matrix.

I understand the equation
(\mathbf{y}-\mathbf{X}\tilde{\beta})^{\text{T}}(\mathbf{y}-\mathbf{X}\tilde{\beta})= \mathbf{y}^{\text{T}}\mathbf{y}-\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{y}-\mathbf{y}^{\text{T}}\mathbf{X}\tilde{\beta}+ \tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{X}\tilde{\beta}<br />
, but then it is stated
\mathbf{y}^{\text{T}}\mathbf{y}-\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{y}-\mathbf{y}^{\text{T}}\mathbf{X}\tilde{\beta}-\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{X}\tilde{\beta}= \mathbf{y}^{\text{T}}\mathbf{y}-2\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{y}+\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{X}\tilde{\beta}\qquad\text{(1)}
, and I do not understand why -\mathbf{y}^{\text{T}}\mathbf{X}\tilde{\beta}=-\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{y} in equation (1).

I understand the transpose identity (\mathbf{y}^{\text{T}}\tilde{\beta}\mathbf{X})^{\text{T}}= \mathbf{X}^{\text{T}}\tilde{\beta}^{\text{T}}\mathbf{y},
but then (1) would be
\mathbf{y}^{\text{T}}\mathbf{y}-\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{y}-\mathbf{y}^{\text{T}}\mathbf{X}\tilde{\beta}-\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{X}\tilde{\beta}= \mathbf{y}^{\text{T}}\mathbf{y}-\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{y}-(\tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{y})^{\text{T}}+ \tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{X}\tilde{\beta},
and (1) would only be true if \tilde{\beta}^{\text{T}}\mathbf{X}^{\text{T}}\mathbf{y} is s symmetric matrix, which I think it need not be.

What am I missing here?

Thanks in advance,

Ed

 

[chiro]

Hey EdMel.

Hint: Is the quantity a scalar? (If it is then what does this imply about the appropriate transpose?)