• ${A_j} \in { \mathbb{C}^{n \times n}},(j = 0,1,2....m)$
  • ${\rm{P(}}\lambda {\rm{) = }}{{\rm{A}}_m}{\lambda ^m} + .....{A_1}\lambda + {A_0}$ is a matrix polynomial, and $\lambda $ is a complex variable and $det({A_m}) = 0$.

Is this true that number distinct roots of $\det (P(\lambda )) = 0$ less than $n \times m$?


Since we are working over $\mathbb{C}$, we can select a basis in which $A_m$ is upper-triangularizable. So, $(A_m)_{j,k} = 0$ for $j >k$, and the determinant is the sum of the product of the diagonals $\sum_{j=1}^n \left( \prod_{k=1}^n (P(\lambda))_{k, j+k-1} - \prod_{k=1}^n (P(\lambda))_{k, j-k-1}\right)$ (This is the "sum of diagonals" method of finding the determinant, and if the indices ever increase over n or below 1 they wrap around in the natural way.)

So, the only place where the polynomial COULD have a $\lambda^{mn}$ term is on the product of the main diagonal. every other diagonal includes some element where $A_m$ is zero, so the polynomial in that spot in $P(\lambda)$ has degree at most $m-1$.

On the main diagonal, the product equals $\prod_{k=1}^n( (A_m)_{k,k} \lambda^m + \dots + (A_0)_{k,k})$. But the leading term of this is $ \left(\prod_{k=1}^n (A_m)_{kk}\right)\lambda^{mn} = Det(A_m) \lambda^{mn}=0$, so the degree of the polynomial is at most $nm-1$.

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  • $\begingroup$ @ JHalliday --Thanks,now let $\det ({A_m}) \ne 0$ is this true that number distinct roots of $\det (P(\lambda )) = 0$ exactly $n \times m$? $\endgroup$ – H.S Sep 11 '15 at 7:44
  • $\begingroup$ Yes that is true by the same argument. The leading term is now nonzero. $\endgroup$ – JHalliday Sep 11 '15 at 7:46
  • $\begingroup$ @ JHalliday -- please send your email Address? $\endgroup$ – H.S Sep 11 '15 at 7:47
  • 1
    $\begingroup$ Um I don't want to, but if you have more questions you can ask me them here. $\endgroup$ – JHalliday Sep 11 '15 at 7:52
  • $\begingroup$ Please check this link (math.stackexchange.com/questions/1383030/…) and please answer $\endgroup$ – H.S Sep 11 '15 at 7:57

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