Let A be a triangular blocks matrix (the blocks are: A1,...,Ak). Show that CA(t)=CA1(t)*...*CAk(t).
Any help ? thanks ;)
(edit: CA and CAj are the characteristic polynomials of the blocks)
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I don't get what $A$ is. Are you sure $A$ is supposed to be triangular? – Git Gud Jun 13 '14 at 15:57
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If A is triangular, the proof is even simpler. But as long as it is block diagonal the statement works. – Mark Fischler Jun 13 '14 at 20:15
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$$C(A)\equiv \det(A - tI)$$ Start from the theorem that determinant is multiplicative $$ \det(KJ) = \det(K) \det(K)$$ Now consider any pair of 2-square-blocks square matrices $$ K ^{(N\times N)}= \left( \begin{array}{cc} k^{r\times r} & 0^{r \times s} \\ 0^{s \times r} & I^{s \times s} \end{array} \right) $$ and $$ J ^{(N\times N)}= \left( \begin{array}{cc} I^{r\times r} & 0^{r \times s} \\ 0^{s \times r} & j^{s \times s} \end{array} \right) $$ It is easy to see that $$ KJ = JK = \left( \begin{array}{cc} k^{r\times r} & 0^{r \times s} \\ 0^{s \times r} & j^{s \times s} \end{array} \right) $$ By induction, it is then clear that the block diagonal matrix $A$ is product of matrices each containing one of the blocks of $A$ and 1's on the remainder of the diagonal. And $A-tI$ has that same structure $$ A - tI = \prod_{i=1}^{k} \mbox{Block}_i (A_k - tI) $$ and $$ C(A) = \det(A - tI) = \prod_{i=1}^{k} \det \left(\mbox{Block}_i (A_k - tI) \right) $$
But the determinant of each of the $mbox{Block}_i (A_k - tI) = \det((A_k - tI)$. And $\det((A_k - tI)$ is C(A_k). So $$ C(A) = \prod_{i=1}^{k} C(A_k)$$
Mark Fischler
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