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Consider a linear operator $T: V \to V$ on a finite-dimensional vector space $V$. Suppose that $U$ is a $T$-invariant subspace of $V$, and consider the map $T/U: V/U \to V/U$ given by $T/U(v + U) = T(v) + U$ (here $V/U$ is the quotient space). I'm interested in how exactly the characteristic polynomials of the three operators $T/U, T,$ and $T|_U$ are related (here $T|_U$ denotes the operator $T$ restricted to $U$). I can show that the eigenvalues of $T|_U$ and the eigenvalues of $T/U$ are also eigenvalues of $T$, so my conjecture is that $p_T(\lambda) = p_{T|_U}(\lambda) \cdot p_{T/U}(\lambda)$, where in general $p_\ell(\lambda)$ is the characteristic polynomial of the linear operator $\ell$. However, I'm not sure how to go about showing that the multiplicities of each eigenvalue "match up".

Binocular10
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1 Answers1

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Choose a basis of $U$, and complete it into a basis of $V$. On the other hand, the vectors that you used to complete the basis can be used to form a basis of $V/U$.

The matrix of $T$ in the completed basis then looks like $$\left[\matrix{M_{T_{U}} &*\\0&M_{T/U}}\right]$$ where $M_{T/U}$ is with respect to the basis from the remark above, and $*$ stands for irrelevant (though possibly non-zero) entries.

Your conjecture follows immediately.

Arnaud Mortier
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  • Does the * in the upper right entry just mean "don't care"? – Binocular10 Apr 28 '18 at 23:43
  • @Binocular10 Yes. – Arnaud Mortier Apr 28 '18 at 23:44
  • So if I understand correctly, the span of the vectors used to complete the basis (of $V$) is the set of coset representatives for each coset of $V/U$? – Binocular10 Apr 28 '18 at 23:53
  • @Binocular10 Exactly. By fixing a set of representatives of a basis of $V/U$, we implicitly fixed a preferred representative for each coset. But note that this remark is not necessary for your purposes, all you need is a basis, and the matrix of $T/U$ in that basis. – Arnaud Mortier Apr 29 '18 at 00:01
  • One thing I'm having trouble with: suppose that $X$ is the set of the vectors used to complete the basis of $V$. Isn't it possible that span$(X)$ contains two coset representatives that belong to the same coset (of $V/U$)? Would this be a problem? In that case would we need to "pare down" the set $X$? But if we do that there's a possibility that the resulting set does not generate all the vectors in $V - U$, which is exactly what we want. – Binocular10 Apr 29 '18 at 00:30
  • @Binocular10 if this happens then the difference between these vectors lies in $U$, by definition of lying in the same coset. But span(X) is in direct sum with $U$, by construction, so the difference is then $0$ and the vectors are actually equal. So no, it isn't possible. – Arnaud Mortier Apr 29 '18 at 00:32
  • Ah I see. Thanks for your help. – Binocular10 Apr 29 '18 at 00:35
  • @Binocular10 You're welcome. – Arnaud Mortier Apr 29 '18 at 00:39