$T:V\to V$ is a linear map on n-dimensional vector space. Suppose $c(x)=(x-d_1)^{e_1}...(x-d_k)^{e_k}$ be its characteristic polynomial where $e_1+e_2+...+e_k=n$. Let $V_1$ be the 1-dimensional subspace generated by an eigenvector corresponding to $d_1$. We consider the natural linear map $T':V/V_1 \to V/V_1$ due to $T$. I am trying to figure out the characteristic polynomial $c'(x)$ for $T'$. I can see that each $d_i$ ,except possibily $d_1$ , will be the root of $c'(x)$ but cannot find its multiplicity. Also the degree of $c'(x)$ will be $n-1$.

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    $\begingroup$ We will simply have $$ c'(x) = \frac {1}{x - d_1} c(x) $$ it is difficult to show this without one of the "powerful" results of linear algebra. Do you know anything about upper-triangualrizing transformations? Something about Jordan form, perhaps? $\endgroup$ – Omnomnomnom Nov 21 '16 at 19:26
  • $\begingroup$ Well this is the result they have used in proof of upper triangularization which I am stuck at. $\endgroup$ – NewB Nov 21 '16 at 19:30
  • $\begingroup$ Have they established any properties of the determinant yet? Do they define the characteristic polynomial in terms of the determinant? $\endgroup$ – Omnomnomnom Nov 21 '16 at 19:34
  • $\begingroup$ yes I am familiar with basic results except canonical forms. $\endgroup$ – NewB Nov 21 '16 at 19:40
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    $\begingroup$ Do you understand what I mean by $V = V/V_1 \oplus V_1$? Or $\det(T_1 \oplus T_2) = \det(T_1) \det(T_2)$? $\endgroup$ – Omnomnomnom Nov 21 '16 at 19:41

Select a basis $\mathcal B = \{v_1,\dots,v_n\}$ where $v_1$ is the eigenvector corresponding to the eigenvalue $d_1$. The matrix of $T$ with respect to $\mathcal B$ has the form $$ [T]_{\mathcal B} = \pmatrix{d_1 & c^T\\&A} $$ where $A$ is an $(n-1)\times (n-1)$ matrix. We then note that $$ \det(xI - T) = \pmatrix{x - d_1 & c^T\\& xI - A} = (x-d_1)\det(x I - A) $$ However, note that $A$ is the matrix of the natural linear map $T'$ with respect to the basis $\{v_2,\dots,v_n\}$.

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  • $\begingroup$ Here's what I think they're going for. What I'm missing here, however, is how exactly do they define the "natural linear map" $T'$? $\endgroup$ – Omnomnomnom Nov 21 '16 at 19:50
  • $\begingroup$ T'(v)=T(v) +V_1 $\endgroup$ – NewB Nov 21 '16 at 20:00
  • $\begingroup$ Ah, cool, then yes this was exactly the right trick $\endgroup$ – Omnomnomnom Nov 21 '16 at 20:11

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