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It is known that all $B$, $C$ and $D$ are $3 \times 3$ matrices. And the eigenvalues of $B$ are $1, 2, 3$; $C$ are $4, 5, 6$; and $D$ are $7, 8, 9$. What are the eigenvalues of the $6 \times 6$ matrix $$\begin{pmatrix} B & C\\0 & D \end{pmatrix}$$ where $0$ is the $3 \times 3$ matrix whose entries are all $0$. From my intuition, I think the eigenvalues of the new $6 \times 6$ matrix are the eigenvalues of $B$ and $D$. But how can I show that?

Travis Willse
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Scorpio19891119
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5 Answers5

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By definition, an eigenvalue $\lambda$ of the block matrix $A$ satisfies

$$\det \begin{pmatrix} B-\lambda I & C \\ 0 & D-\lambda I \end{pmatrix} = 0.$$

Using a property of block matrix determinants, we have

$$\det \begin{pmatrix} B-\lambda I & C \\ 0 & D-\lambda I \end{pmatrix} = \det(B-\lambda I)\det(D-\lambda I) = 0$$

Thus the eigenvalues of $B,D$ are also the eigenvalues of $A$.

Jacob
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If $\lambda$ is an eigenvalue of $A$ with eigenvector $(x_1, x_2, x_3)^t$ then $(x_1,x_2,x_3,0,0,0)^t$ is an eigenvector of the block matrix. Similarly, for $D$ but you put three zeros at the beggining.

PAD
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    Not true for the eigenvectors of $D$. However the dual eigenvectors (row vectors that are eigenvector for multiplication to the right) of $D$ are (after extension by zeroes) dual eigenvectors of the whole matrix. – Marc van Leeuwen Oct 05 '12 at 19:42
  • You are right! So, there is some use for dual eigenvectors! – PAD Oct 07 '12 at 10:38
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Hint: $$ \pmatrix{B & C\cr 0 & D\cr} \pmatrix{B^{-1} & E\cr 0 & D^{-1}\cr} = \pmatrix{I & BE + CD^{-1}\cr 0 & I\cr} $$ What $E$ will make $BE + CD^{-1} = 0$?

Robert Israel
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  • Would you mind to explain a little more – D. Sikilai Apr 04 '24 at 21:45
  • @D.Sikilai If $B$ and $D$ are invertible, then $E = - B^{-1} C D^{-1}$ works, so $\matrix{B & C\cr 0 & D\cr}$ is invertible. That says that any eigenvalue of $\matrix{B & C\cr 0 & D\cr}$ is an eigenvalue of $B$ or $D$. – Robert Israel Apr 07 '24 at 02:25
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$BE=-CD^{-1}$ By your assumptions $B$ is invertible, hence we have: $-B^{-1}CD^{-1}$

Robert Shore
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hadi
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0

Here is yet another proof. We have that for square matrices $A,C$

$$\begin{pmatrix}A&B \\ O& C\end{pmatrix}=\begin{pmatrix}I&O \\ O& C\end{pmatrix}\begin{pmatrix}I&B \\ O& I\end{pmatrix}\begin{pmatrix}A&O \\ O& I\end{pmatrix}$$

and since the middle matrix is upper triangular, we indeed get

$$\det\begin{pmatrix}A&B \\ O& C\end{pmatrix}=\det(A)\det(C)$$

and so $$\det\begin{pmatrix}A-\lambda I&B \\ O& C-\lambda I\end{pmatrix}=\det(A-\lambda I)\det(C-\lambda I).$$

Thomas Kojar
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