proof of determinant lower bound of a strict diagonally dominant matrix
Let’s define, for any
Then, by strict diagonally dominance, one has . Let and , so that the i-th row of matrix is equal to the corresponding row of matrix multiplied by . In this way , one has
so that, being ,
In this way, we found that each eigenvalue of is greater than one in absolute value; for this reason,
whence the thesis.
Remark: Perhaps it could be not immediately evident where the hypothesis of strict diagonally dominance is employed in this proof; in fact, inequality (*) and (**) would be, in a general case, not valid; they can be stated only because we can assure, by virtue of strict diagonally dominance, that the final argument of the absolute value () does remain positive.
|Title||proof of determinant lower bound of a strict diagonally dominant matrix|
|Date of creation||2013-03-22 17:01:11|
|Last modified on||2013-03-22 17:01:11|
|Owner||Andrea Ambrosio (7332)|
|Last modified by||Andrea Ambrosio (7332)|
|Author||Andrea Ambrosio (7332)|