Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Hi am having trouble with this question

Let $A = \left[\begin{array}{cc}-1&3&-4&-3\\2&-1&0&-3\end{array}\right]$ Find the coordinate vector for the matrix $A \cdot A^{T}$ with respect to the standard basis for $R^{2 \times 2}$

I figured out the $A \cdot A^{T} = \left[\begin{array}{cc}5&-5&-4&-3\\-5&10&12&-6\\-4&12&16&-12\\-3&-6&12&18\end{array}\right]$ and the standard basis for $R^{2 \times 2}$ is $\left[\begin{array}{cc}1&0\\0&0\end{array}\right], \left[\begin{array}{cc}0&1\\0&0\end{array}\right], \left[\begin{array}{cc}0&0\\1&0\end{array}\right], \left[\begin{array}{cc}0&0\\0&1\end{array}\right]$but that is it I get stuck at this point because i cant figure out how $2\times2$ matrices can result to a $4 \times 4$ matrix

share|improve this question
    
You have a mistake: you matrix $\;A\; $ is $\;2\times 4\implies A^t\;$ is $\;4\times 2\;$, and thus their product $\;AA^t\;$ must be $\;2\times 2\;$ ... –  DonAntonio Oct 14 '13 at 4:03

1 Answer 1

up vote 0 down vote accepted

$$AA^t=\begin{pmatrix}-1&\;\;3&-4&-3\\\;\;2&-1&\;\;0&-3\end{pmatrix}\begin{pmatrix}-1&\;\;2\\\;\;3&-1\\-4&\;\;0\\-3&-3\end{pmatrix}=\begin{pmatrix}35&4\\4&14\end{pmatrix}$$

share|improve this answer
    
THANKS A LOT makes sense now I did $A^{t} \times A$ thinking it would be the same –  MATRIXAREHARD Oct 14 '13 at 4:14
    
Nop. $\;A^tA\;$ is a $\;4\times 4\;$ matrix...which is what you calculated, of course. The order in matrix product matters a lot. –  DonAntonio Oct 14 '13 at 4:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.