# Need help with expanding $(\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})$

1. Nov 16, 2011

### devonho

1. The problem statement, all variables and given/known data

I need help with expanding:

$(\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})$

$\bf{x,\mu_{i}}$ are column vectors.
$\Sigma$ is a square matrix.

Thank you.

2. Relevant equations

Can:

$\bf{x}^t\Sigma^{-1}\bf{\mu_{i}}$

be written as?

$\bf{\mu_{i}}^t\Sigma^{-1}\bf{x}$

I've tried to compute this numerically and the answer is no.

3. The attempt at a solution

$(\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}}) = \bf{x}^t\Sigma^{-1}\bf{x}+ \bf{\mu_{i}}^t\Sigma^{-1}\bf{\mu_{i}}- \bf{x}^t\Sigma^{-1}\bf{\mu_{i}}- \bf{\mu_{i}}^t\Sigma^{-1}\bf{x}$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 16, 2011

Re: Need help with expanding $(\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})[/i I don't understand your problem. The result in (3) is correct. If *you* derived that result, then you have obtained the desired result. On the other hand, if you mean that somebody else has given you the result in (3) and you don't know how they got it, that is a different question. So, what, exactly are you asking? RGV 3. Nov 16, 2011 ### Mark44 ### Staff: Mentor Re: Need help with expanding [itex](\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})[/i BTW, [itex]\Sigma$ is a horrible name for a matrix for the reason that it is used primarily to mean summation.

4. Nov 16, 2011

Re: Need help with expanding $(\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})[/i I agree with you. Nevertheless, it is often used in Statistics and Econometrics, etc., to denote the variance-covariance matrix of a multidimensional random variable. RGV 5. Nov 16, 2011 ### statdad Re: Need help with expanding [itex](\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})[/i The answer is yes IF the matrix $$\Sigma$$ is symmetric. Since $$\bf{\mu_i}^t \Sigma^{-1} \bf{x}$$ is a scalar, it equals its transpose, so $$\bf{\mu_i}^t \Sigma^{-1} \bf{x} = \left(\bf{\mu_i}^t \Sigma^{-1} \bf{x}\right)^t = \bf{x}^t \Sigma^{-1} \bf{\mu_i}$$ Did the example you used have sigma symmetric? (If it is a variance-covariance matrix, it has to be symmetric). 6. Nov 16, 2011 ### Ray Vickson Re: Need help with expanding [itex](\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})[/i The way (3) is written, it is valid even if the matrix is not symmetric: the terms x^T A m and m^T A x are written separately. RGV 7. Nov 16, 2011 ### devonho Re: Need help with expanding [itex](\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})[/i Hi all, thanks for the replies. The goal was to get: [itex](\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}}) = \bf{x}^t\Sigma^{-1}\bf{x}+ \bf{\mu_{i}}^t\Sigma^{-1}\bf{\mu_{i}}- 2\bf{x}^t\Sigma^{-1}\bf{\mu_{i}}$

Hence,
$\bf{\mu_i}^t \Sigma^{-1} \bf{x} = \bf{x}^t \Sigma^{-1} \bf{\mu_i}$

Was what I needed. Thanks.

8. Nov 16, 2011

### devonho

Re: Need help with expanding $(\bf{x-\mu_{i}})^{t}\Sigma^{-1}(\bf{x-\mu_{i}})[/i Thanks for the help. I wrote the long proof. If [itex] m_{12}=m_{21}, m_{13}=m_{31},m_{32}=m_{23},$

$\bf{x} = \left[ \begin{array}{ccc} x_1 \\ x_2 \\ x_3 \end{array} \right]$

$\bf{y} = \left[ \begin{array}{ccc} y_1 \\ y_2 \\ y_3 \end{array} \right]$

$\bf{M} = \left[ \begin{array}{ccc} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \end{array} \right]$

then

$\bf{y^tMx}=$
$\begin{array}{ccc} x_1y_1m_{11}+ (x_1y_2+x_2y_1)m_{12}+ (x_1y_3 + x_3y_1)m_{13}+ x_2y_2m_{22}+ (x_2y_3 + x_3y_2)m_{23}+ x_3y_3m_{33} \end{array}$

$\bf{x^tMy}=$
$\begin{array}{ccc} \begin{array}{ccc} (x_1m_{11}+x_2m_{21}+x_3m_{31})y_1 + (x_1m_{12}+x_2m_{22}+x_3m_{32})y_2 + (x_1m_{13}+x_2m_{23}+x_3m_{33})y_3 \end{array} \\= \begin{array}{ccc} x_1y_1m_{11}+ (x_2y_1+ x_1y_2)m_{12}+ (x_3y_1+ x_1y_3)m_{13}+ x_2y_2m_{22}+ (x_3y_2+ x_2y_3)m_{23}+ x_3y_3m_{33} \end{array} \end{array}$
$=\bf{y^tMx}$

Last edited: Nov 16, 2011