1
$\begingroup$

Let $A,B,U,O$ four matrix of real entries. $A$ is a square matrix of size $m$, $B$ is a square matrix of size $n$, while $U$ is a $n\times m$ matrix of all entries $=-1$ and $O$ is a $m\times n$ matrix of all entries $=0$. Let us consider the square (block) matrix of size $n+m$: $$M=\left( \begin{array}{cc} A & U \\ O & B \\ \end{array} \right) $$ Prove that the set of eigenvalues of $M$ is the union of eigenvalues of $A$ and $B$. This property of determinants should follow immediately if we can show $\det M=\det A \det B$ but I do not know to prove this statement. Any suggestions please?

$\endgroup$

marked as duplicate by daw, muaddib, Fly by Night, user147263, Ken Aug 9 '15 at 22:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

Hint: Note that $$ M = \pmatrix{A&O^T\\O&B} \pmatrix{I & A^{-1}U\\0 & I} $$ Now, you just need to show that the first matrix has determinant $\det(A)\det(B)$, and the second has determinant $1$. The second matrix is upper triangular.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.