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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?

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You might want to work on your acceptance rate –  Jacob Oct 5 '12 at 17:51
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math.stackexchange.com/faq#howtoask –  Jacob Oct 5 '12 at 21:08
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3 Answers 3

up vote 6 down vote accepted

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 $A$.

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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 $BE + CD^{-1} = 0$?

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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.

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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 5 '12 at 19:42
    
You are right! So, there is some use for dual eigenvectors! –  PAD Oct 7 '12 at 10:38
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