# Problem of the Week #48 - April 29th, 2013

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Let $V$ be a finite dimensional complex vector space. Let $\phi$ be an element of $\text{End}_{\mathbb{C}}(V)$, and consider the function $f:\mathbb{C}\rightarrow\mathbb{C}$ by $f(z)=\det(1+z\cdot\phi).$
Find an expression for $f^{\prime}(0)$ using what is known about trace, determinants, and the characteristic polynomial.

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#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. You can find my solution below.

Recall that we define the characteristic polynomial as$P_{\text{char}}^{\phi} = \det(x\cdot\mathbf{1}-\phi).$
Now, we know that the eigenvalues of $\phi$ are exactly the roots of $P_{\text{char}}^{\phi}$. Take $\lambda=-\frac{1}{z}$. Then
\begin{aligned}\det(\mathbf{1}+z\cdot\phi) &= z^n\det(\frac{1}{z}+\phi)\\ &= z^n\det(\phi-\lambda\cdot\mathbf{1})\\ &=z^n(a_1-\lambda)(a_2-\lambda)\cdots(a_n-\lambda)\\ &=(1+za_1)(1+za_2)\cdots(1+za_n)\end{aligned}
where $a_1,\ldots,a_n$ are the eigenvalues of $\phi$. Therefore,
$f^{\prime}(z) = a_1(1+za_2)\cdots(1+za_n)+a_2(1+za_1)(1+za_3)\cdots(1+za_n)+\cdots+a_n(1+za_1)\cdots(1+za_{n-1}).$
Thus, $f^{\prime}(0)=a_1+a_2+\cdots+a_n=\text{tr}(\phi)$.

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