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[SOLVED] Exterior Power of Linear Transformation

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi everyone, :)

I don't understand how to use the given linear transformation so as to calculate the exterior power of \(V\); \(\wedge^2(f)\). I hope you can help me with this. :)

Problem:

Find the trace of the linear transformation \(\wedge^2(f)\), if \(f\) is given by the matrix,

\[A=\begin{pmatrix}1&1&0\\0&2&2\\0&0&3\end{pmatrix}\]
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi everyone, :)

I don't understand how to use the given linear transformation so as to calculate the exterior power of \(V\); \(\wedge^2(f)\). I hope you can help me with this. :)

Problem:

Find the trace of the linear transformation \(\wedge^2(f)\), if \(f\) is given by the matrix,

\[A=\begin{pmatrix}1&1&0\\0&2&2\\0&0&3\end{pmatrix}\]
I think I got this one covered. Let me know if you see any mistakes. :)

For a \(n\times n\) matrix \(A\) the characteristic polynomial is given by the general formula,

\[P_{A}(\lambda)=\sum_{k=0}^{n}\lambda^{n-k}(-1)^{n-k}\mbox{tr}(\wedge^k A)\]

Therefore for a \(3\times 3\) matrix this reduces to,

\[P_{A}(\lambda)=-\lambda^3+\mbox{tr}(\wedge A)\lambda^2-\mbox{tr}(\wedge^2 A)\lambda+\mbox{tr}(\wedge^3 A)\]

Now if we write the characteristic polynomial of \(A\) we get,

\[P_{A}(\lambda)=(1-\lambda)(2-\lambda)(3-\lambda)=-\lambda^2+6\lambda^2-11\lambda+6\]

Therefore,

\[\mbox{tr}(\wedge^2 A)=11\]
 
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