Unformatted text preview: hat is not upper triangular.
If it were upper triangular, backsubstitution would imply that x 0 . Hence a linear
combination of all the rows results in a row containing only zeros. That is, there are
scalars, , one for each row and not all zero, such that for each for column ,
Now consider A
for each , . By definition, , or It follows that .
Let y . We therefore have nonzero vector, y , such that A y 0. 26. By definition,
Ax y Let , so that Ax Now interchanging the order or summation,
Ax y Now note that ________________________________________________________________________
page 359 —————————————————————————— CHAPTER 7. —— Ay .
Ax y Ay x Ay . 28. By linearity,
Ax 29. Let Ax
b A . By the hypothesis, there is a nonzero vector, y , such that
, . Taking the conjugate of both sides, and interchanging the indices, we have This implies that a linear combination of each row of A is equal to zero. Now consider
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- Spring '08
- Linear Algebra, eigenvector