What are the eigenvalues of following block matrix?
$$\begin{bmatrix} A & B \\ B^T & O \end{bmatrix}$$
Here, $A$ and $B$ are any square matrices of order $n$, $O$ is zero matrix of order $n$.
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asked Apr 6, 2016 at 11:09
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2$\begingroup$ Since $B^T$ and $\lambda I$ commute we have the formula below: $$ P(\lambda) = \text{det} \left[ \begin{array}{cc} \lambda I - A & -B \\ -B^T & \lambda I \end{array} \right] = \text{det}\left[ (\lambda I - A)(\lambda I)-BB^T \right]$$ thus, $$ P(\lambda) = \text{det}\left[ \lambda^2 I - \lambda A -BB^T \right] $$ Let $M = \lambda^2 I - \lambda A -BB^T$... $\endgroup$– James S. CookCommented Apr 6, 2016 at 11:31
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2$\begingroup$ ... Notice $BB^T$ is symmetric and real (I assume, I suppose you can generalize over $\mathbb{C}$ with general spectral theorem) thus diagonalizable via an orthonormal eigenbasis with matrix $P$ such that $P^TP=I$. In particular $D_B = P^T(BB^T)P$ is diagonal. Moreover, $$ P^TMP = (\lambda^2I+D_B)-\lambda P^TAP $$ (this is not an answer, but, perhaps someone can make use of it..) $\endgroup$– James S. CookCommented Apr 6, 2016 at 11:32
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1 Answer
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Generally, determinant of block matrix can be obtained following formula.
$$ \det \begin{bmatrix} A_{0} & B_{0} \\ C_{0} & D_{0} \end{bmatrix} = \det(D_{0}) \det(A_{0}-B_{0}D_{0}^{-1}C_{0}) $$
Then, this problem's case: $(A_{0},B_{0},C_{0},D_{0})=(A-\lambda E,B,B^{\text{T}},-\lambda E)$, where $\lambda$ is eigenvalue. Therefore,
$$ \begin{align} \det \begin{bmatrix} A-\lambda E & B \\ B^{\text{T}} & -\lambda E \end{bmatrix} =& \det(-\lambda E) \det((A-\lambda E)-B(-\lambda E)^{-1}B^{\text{T}}) \\ =& \det(-\lambda(A-\lambda E)-B B^{\text{T}}) \\ =& \det(\lambda^2 E -\lambda A -B B^{\text{T}}) \end{align} $$
This determinant only cannot identify the eigenvalues as far as $A$ and $B$ are not supplied concrete elements. However, we can get ingredients of matrices how this quadratic matrix polynomial is made to decompose. To sum up, we can see what eigenvalues are constructed by $n$ orders matrices. Hence, we attempt to factorize above determinant.
$$ \begin{align} \det(\lambda^2 E -\lambda A -B B^{\text{T}})=&\det(X-\lambda E)(Y - \lambda E) \\ =&\det(X-\lambda E)\det(Y - \lambda E) \\ \end{align} $$
Above determinants are changeable so that following conditions can be applied to solve these matrices as $X,Y$. Especially, secondly condition is allowed commutative.
$$ \begin{cases} X+Y=A \\ XY=YX=-BB^{\text{T}} \end{cases} $$
Because,
$$ \begin{cases} \det(X-\lambda E)(Y - \lambda E)=\det(\lambda^2 E-\lambda(X+Y)+XY) \\ \\ \det(Y-\lambda E)(X - \lambda E)=\det(\lambda^2 E-\lambda(X+Y)+YX) \end{cases} $$
Thus, $X,Y$ are estimated below.
$$ \det\biggl(\cfrac{A+\sqrt{A^2+4BB^{\text{T}}}}{2}-\lambda E\biggr) \det\biggl(\cfrac{A-\sqrt{A^2+4BB^{\text{T}}}}{2}-\lambda E\biggr) =0 $$
In conclusion, eigenvalues $\lambda$ show these matrices. Its means that the block matrix includes $X$'s eigenvalues $\lambda_{X}$ and $Y$'s eigenvalues $\lambda_{Y}$. Incidentally, $A=\pm 2i\sqrt{BB^{\text{T}}}$ are special cases owing to multiple root as $X=Y$. Then, eigenvalues $\lambda$ correspond to just half $A$'s eigenvalues $\lambda_{0.5A}$.
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$\begingroup$ I don't see why $XY=YX$ when $X,Y$ are $\frac{A\pm\sqrt{A^2-4BB^T}}{2}$. If $X$ and $Y$ commute, then $A$ commutes with $\sqrt{A^2-4BB^T}$ and in turn with $A^2-4BB^T$ and $BB^T$. But this clearly doesn't hold in general because $A$ and $B$ are arbitrary.. $\endgroup$– user1551Commented Apr 15, 2016 at 4:11
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$\begingroup$ Absolutely. my explanation was defected. I have editted why $XY$ is changeable. $\endgroup$ Commented Apr 15, 2016 at 4:22
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$\begingroup$ No. You have only stated the obvious fact that $\det[(X-\lambda E)(Y-\lambda E)]=\det[(Y-\lambda E)(X-\lambda E)]$, but you haven't explained why $XY\ne YX$ or why $\det(\lambda^2 E-\lambda A-BB^T)=\det[(X-\lambda E)(Y-\lambda E)]$ $\endgroup$– user1551Commented Apr 15, 2016 at 4:40
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$\begingroup$ Why I factorize the determinant. I guess concrete eigenvalues cannot be obtained by only $A$ and $B$. But, this is a possible attempt what eigenvalues are composed by $n$-orders matrices. I think this is good clue. $\endgroup$ Commented Apr 15, 2016 at 6:40