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Unformatted text preview: matrix A b . Replace the last row by We find that if b y
, then the last row of the augmented matrix contains only zeros.
Hence there are
remaining equations. We can now set
, some parameter,
and solve for the other variables in terms of
Therefore the system of equations
Ax b has a solution.
30. If is an eigenvalue of A , then there is a nonzero vector, x , such that
Ax x 0. That is, Ax 0 has a nonzero solution. This implies that the mapping defined by A is
not 1-to-1, and hence not invertible. On the other hand, if A is singular, then
Thus, Ax 0 has a nonzero solution. The latter equation can be written as Ax
31. As shown in Prob. , Ax y x A y . By definition of a Hermitian matrix, ________________________________________________________________________
page 360 —————————————————————————— CHAPTER 7. ——
A A 32 . Based on Prob. , Ax x x Ax . . Let x be an eigenvector corresponding to an eigenvalue
It then follows that
x x and x Ax
x x Based on the properties of the inner product,
x x and x x
x x . Then from Part
. From Part xx ,
xx Based on the defin...
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This note was uploaded on 03/11/2014 for the course MA 303 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.
- Spring '08