Page 165 - DMTH502_LINEAR_ALGEBRA

Unit 12: Introduction and Characteristic Values of Elementary Canonical Forms

On solving these equations, we get Notes
x 1 x 2 x 3 k (say)
1 2 2 1
Hence the required characteristic vector corresponding to the characteristic root , = 0, is
1
x 1 k 1
X x 2 2k 1
x 2k
3 1
The characteristic vector corresponding to the root = 3 is given by
2
8 3 6 2 x 0
1
6 7 3 4 x 2 0
2 4 3 3 x 3 0

5 6 2 x 1 0
6 4 4 x 0
or 2
2 4 0 x 0
3
This gives 5x – 6x + 2x = 0
1 2 3
6x + 4x – 4x = 0
1 2 3
2x – 4x = 0
1 2
On solving these equations, we get
x 1 x 2 x 3 k (say) k 0
2 1 2 2 2
x 1 2k 2
Thus x = x 2 k 2 is the required characteristic vector for = 3.
x 3 2k 2
Similarly, for = 15, the characteristic vector will be

8 15 6 2 x 0
1
6 7 15 4 x 2 0
2 43 3 15 x 3 0

7 6 2 x 1 0
6 8 4 x 0
or 2
2 4 12 x 3 0
which give 7x + 6x – 2x = 0
1 2 3
3x + 4x + 2x = 0
1 2 3
x – 2x – 6x = 0
1 2 3
On solving these, we get
x 1 x 2 x 3 k (say) k 0
2 2 1 3 3

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