Page 164 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Suppose that T is the linear operator on R which is represented by A in the standard basis. Let
3
us find the characteristic vectors of T associated with the characteristic values, 1 and 2. Now
2 1 1
A I 2 1 1
2 2 1

It is obvious at a glance that A–I has rank equal to 2 (and hence T – I has nullity equal to 1). So the
space of characteristic vectors associated with the characteristic value 1 is one-dimensional. The
vector = (1, 0, 2) spans the null space of T – I . Thus T = if and only if is a scalar multiple
1
of . Now consider
1
1 1 1
A 2I 2 0 1
2 2 2

Evidently A – 2I also has rank 2, so that the space of characteristic vectors associated with value
2 has dimension 1. T = 2 is possible if is a scalar multiple of = (1, 1, 2).
2

Example 3: Find the characteristic values and associated characteristic vector for the
matrix
8 6 2
A 6 7 4
2 4 3
Solution: We know that the characteristic equation is |A – I| = 0, i.e.,

8 6 2
6 7 4 0
2 4 3

or {(8 – )} (7 – ) (3 – ) – 16} + 6{3 – } (–6) + 8} + 2 {24 – 2(7 – )} = 0
2
or – + 18 – 45 = 0
3
2
or ( + 18 + 45) = 0
or ( – 3) ( – 15) = 0
= 0, 3, 15.
Hence the characteristic roots are = 0, = 3, = 15. The characteristic vector associated with
1 2 3
is = 0 is given by
1
8 6 2 x 1 0
6 7 4 x 0
2
2 4 3 x 3 0
This gives 8x – 6x + 2x = 0
1 2 3
–6x + 7x – 4x = 0
1 2 3
2x – 4x + 3x = 0
1 2 3

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