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Unformatted text preview: 18.06  Spring 2005  Problem Set 8
Solution to the Challenge Problem Challenge Problem: Consider the 3 × 3 matrix
�
�
abc
A = ⎞1 d e �
01f
Determine the entries a, b, c, d, e, f so that:
• the top left 1 × 1 block is a matrix with eigenvalue 2;
• the top left 2 × 2 block is a matrix with eigenvalues 3 and 3;
• the top left 3 × 3 block is a matrix with eigenvalues 0, 1 and 2.
Solution. Let Ai denote the top left i × i block of A. The matrix A1 is the
matrix (a). Since a is the only eigenvalue of this matrix, we conclude that a = 2.
We now move on to determining the entries of the matrix A2 , the top left
2 × 2 block of A:
⎠
⎛
2b
A2 =
1d
Since the sum of the eigenvalues of A2 is 0 by hypothesis, and it is also equal
to the trace of A2 , we obtain that 2 + d = 0, or d = −2. Moreover, the product
of the eigenvalues of A2 is 9 by hypothesis, and it is equal to the determinant
of A2 . Thus we have
−9 = 2d − b = −4 − b
and we deduce that b = 5 and therefore
⎠
⎛
25
A2 =
1 −2
Finally, consider A = A3 . Again, the sum of the eigenvalues of A is 1 and
it is also equal to the trace of A. We deduce that f = −1. We still need to
determine the entries c and e of A, and we have
�
�
25
c
A = ⎞ 1 −2 e �
0 1 −1
The characteristic polynomial of this matrix is
−�3 − �2 + (e + 9)� + c − 2e + 9 1 We know that the roots of this polynomial must be 0, 1 and 2. Setting
� = 0 and � = 1 we obtain
c − 2e + 9 = 0
−1 − 1 + (e + 9) + c − 2e + 9 = 0
which are equivalent to
c − 2e = −9
c − e = −16
Thus c = −7 and e = 9 and we conclude
�
�
2 5 −7
A = ⎞ 1 −2 −9 �
0 1 −1 2 ...
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This note was uploaded on 11/16/2011 for the course MATH 201 taught by Professor Soysal during the Fall '08 term at Boğaziçi University.
 Fall '08
 SOYSAL

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