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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

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  • $\begingroup$ 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 $\;2\times 2\;$ ... $\endgroup$
    – DonAntonio
    Oct 14 '13 at 4:03
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$$AA^t=\begin{pmatrix}-1&\;\;3&-4&-3\\\;\;2&-1&\;\;0&-3\end{pmatrix}\begin{pmatrix}-1&\;\;2\\\;\;3&-1\\-4&\;\;0\\-3&-3\end{pmatrix}=\begin{pmatrix}35&4\\4&14\end{pmatrix}$$

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

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