# [SOLVED]Exterior Power of Linear Transformation

#### Sudharaka

##### Well-known member
MHB Math Helper
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
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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