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I'm recently playing with some $(n+k)\times (n+k)$ block matrices: \begin{equation*} \mathbf{X} = \begin{pmatrix} \mathbf{0}_{n\times n} & \mathbf{P}_{n\times k} \\ {\mathbf{P}}_{k\times n}^T & \mathbf{0}_{k\times k} \end{pmatrix} \end{equation*} where $\mathbf{P}_{n\times k}$ consists of $k$ orthonormal basis in $\mathbb{R}^n$ so that $\mathbf{P}^T\mathbf{P} = \mathbf{I}_{k\times k}$. After playing with Matlab, I conjectured that $\mathbf{X}$ has a very special set of eigenvalues. \begin{equation*} \text{Eigenvalues of $\mathbf X$} = \text{$k$ copies of -1, $k$ copies of +1, and $n-k$ copies of 0.} \end{equation*} There can be many ways to prove this, but I tried to compute the characteristic polynomial $p_{\mathbf{X}}(\lambda)$ of $\mathbf{X}$

\begin{equation*} p_{\mathbf{X}}(\lambda) = \det(\lambda\mathbf{I} - \mathbf{X}) = \det \begin{pmatrix} \lambda\mathbf{I}_n & -\mathbf{P} \\ -\mathbf{P}^T & \lambda\mathbf{I}_k \end{pmatrix} \end{equation*}

to show that $p_{\mathbf{X}}(\lambda) = (\lambda^2-1)^k\lambda^{n-k}$ which will prove my conjecture because the roots of a characteristic polynomial $p_{\mathbf{X}}(\lambda)$ are the eigenvalues of $\mathbf{X}$.

Maybe I can exploit Schur determinant identity which states that \begin{equation*} \det \begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{pmatrix} = \det\mathbf{A}\cdot \det\left( \mathbf{D} - \mathbf{C}\mathbf{A}^{-1}\mathbf{B} \right) \end{equation*} assuming $\mathbf A$ is invertible. Well, if I can assume $\lambda \neq 0$, then I can easily compute the characteristic polynomial using this identity. \begin{align*} \det \begin{pmatrix} \lambda\mathbf{I}_n & -\mathbf{P} \\ -\mathbf{P}^T & \lambda\mathbf{I}_k \end{pmatrix} &= \det(\lambda\mathbf{I}_n)\cdot \det\left( \lambda\mathbf{I}_k - \mathbf{P}^T(\lambda\mathbf{I}_n)^{-1}\mathbf{P} \right) \\ &= \det(\lambda\mathbf{I}_n)\cdot\det \left[\left(\lambda - \frac{1}{\lambda}\right)\mathbf{I}_k\right] \\ &= \lambda^n\left(\lambda - \frac{1}{\lambda}\right)^k \\ &= \lambda^{n-k}(\lambda^2 - 1)^k \end{align*}

  1. This seems the right answer but it only holds when $\lambda\neq 0$. How can I show that this indeed holds even when $\lambda = 0$? Maybe I can try the cofactor expansion for determinant but that seems not easy for this problem.

  2. I once heard some people saying "the invertible matrices are dense in the space of all matrices, and we can conclude the general case holds as well by the continuity argument" in their papers but I cannot truly understand what they're saying. Can I just repeat what they said for my proof?

  3. This is just an additional question but where can I find some explanation on such a continuity argument?

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    $\begingroup$ No continuity needed. If two polynomials with real coefficients are equal to each other at infinitely many points (e.g., at all nonzero reals, in your case), then they are identical. See math.uconn.edu/~kconrad/blurbs/linmultialg/univid.pdf and math.uconn.edu/~kconrad/blurbs/linmultialg/univid2.pdf for a thorough discussion of this tactic and similar ones. $\endgroup$ – darij grinberg Jul 19 '15 at 13:33
  • $\begingroup$ @darijgrinberg: I'm not sure whether I can say two polynomials are equal to each other at infinitely many points. Which two polynomials do I have in this problem? I computed one characteristic polynomial assuming $\lambda\neq 0$, but what's the other? $\endgroup$ – user19906 Jul 19 '15 at 13:42
  • $\begingroup$ The one polynomial is the characteristic polynomial; the other is $\lambda^{n-k}\left(\lambda^2-1\right)^k$. $\endgroup$ – darij grinberg Jul 19 '15 at 13:49
  • $\begingroup$ @darijgrinberg: I'm sorry but I still don't understand. You're saying that there's an way to evaluate $p_{\mathbf{X}}(\lambda) = \det\begin{pmatrix}\lambda\mathbf{I}_n & \mathbf{P} \\ \mathbf{P}^T & \lambda\mathbf{I}_k\end{pmatrix}$ and compare it to $\lambda^{n-k}(\lambda^2-1)^k$ to see whether these two are evaluated to the same value when $\lambda\neq 0$? $\endgroup$ – user19906 Jul 19 '15 at 14:35
  • $\begingroup$ @user19906 Isn't that what you've done in your post above after words "Well, if I can assume $\lambda\ne 0$, then..."? $\endgroup$ – A.Γ. Jul 19 '15 at 14:52

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