My Math Forum determinant of BA.....

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 November 13th, 2011, 04:25 AM #1 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus determinant of BA..... [color=#000000]Let $A\in \mathbb{M}_{3,2}(\mathbb{C})$ and $B\in\mathbb{M}_{2,3}(\mathbb{C})$ such that $\hspace{400pt}AB=\left(\begin{matrix} 1 &-1=&2 \\0=&2\\0=&2\end{matrix}\right)=$ what is the value of $\det(BA)$?[/color]
 November 16th, 2011, 10:54 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: determinant of BA..... Assuming there is enough information to solve this, then the precise A and B are not important. So just find A and B that work: in particular $\begin{bmatrix}1=&0 \\ 0=&1 \\ 0=&1\end{bmatrix}\begin{bmatrix}1=&-1=&2 \\ 0=&2\end{bmatrix}= \begin{bmatrix}1 &-1=&2 \\ 0=&2 \\ 0=&2\end{bmatrix}=$ Calculate BA with those A and B and find its determinant.
 November 16th, 2011, 11:21 AM #3 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: determinant of BA..... 2???
 November 17th, 2011, 12:49 AM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: determinant of BA..... [color=#000000]Yes wnvl! The eigenvalues of the matrix AB are 0,1,2, so the characteristic polynomial of AB is $\hspace{350pt}p_{AB}(\lambda)=\lambda(\lambda-1)(\lambda-2)=\lambda^3-3\lambda^2+2\lambda$ so the characteristic polynomial of BA is $\hspace{350pt}p_{BA}(\lambda)=\lambda^{2-3}p_{AB}(\lambda)=\lambda^2-3\lambda+2$ meaning that det(BA)=2.[/color]
November 17th, 2011, 12:59 AM   #5
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Re: determinant of BA.....

Quote:
 Originally Posted by ZardoZ [color=#000000]Yes wnvl! The eigenvalues of the matrix AB are 0,1,2, so the characteristic polynomial of AB is $\hspace{350pt}p_{AB}(\lambda)=\lambda(\lambda-1)(\lambda-2)=\lambda^3-3\lambda^2+2\lambda$ so the characteristic polynomial of BA is $\hspace{350pt}p_{BA}(\lambda)=\lambda^{2-3}p_{AB}(\lambda)=\lambda^2-3\lambda+2$ meaning that det(BA)=2.[/color]
Just said 2 after reading the post of HallsofIvy.

$p_{BA}(\lambda)=\lambda^{size(BA)-size(AB)}p_{AB}(\lambda)$

Didn't know that theorema, but not unlogic.

 November 17th, 2011, 01:20 AM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: determinant of BA..... [color=#000000]Yes but HallsofIvy made an assumption that might have led to "questionable" conclusions.[/color]

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