Page 170 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes x 5 6
= (x 2)
2 x 2
2
= (x – 2) (x – 3x + 2)
2
= (x – 2) (x – 1).
What are the dimensions of the spaces of characteristic vectors associated with the two
characteristic values? We have

4 6 6
A – I = 1 3 2
3 6 5

3 6 6
A – 2I = 1 2 2
3 6 6

We know that A – I is singular and obviously rank (A – I) 2. Therefore, rank (A – I) = 2. It is
evident that rank (A – 2I) = 1.

Let W , W be the spaces of characteristic vectors associated with the characteristic values 1, 2. We
1 2
know that dim W = 1 and dim W = 2. By Theorem 2, T is diagonalizable. It is easy to exhibit a
1 2
3
basis for R in which T is represented by a diagonal matrix. The null space of (T – I) is spanned by
the vector = (3, –1, 3) and so { } is a basis for W . The null space of T – 2I (i.e., the space W )
1 1 1 2
consists of the vectors (x , x , x ) with x = 2x + 2x . Thus, one example of a basis for W is
1 2 3 1 2 3 2
= (2, 1, 0)
2
= (2, 0, 1).
3
If = ( , , ), then [T] is the diagonal matrix
1 2 3
1 0 0
D = 0 2 0
0 0 2

The fact that T is diagonalizable means that the original matrix A is similar (over R) to the
diagonal matrix D. The matrix P which enables us to change coordinates from the basis to the
standard basis is (of course) the matrix which has the transposes of , , as its column
1 2 3
vectors:
3 2 2
P = 1 1 0
3 0 1

Furthermore, AP = PD, so that
–1
P AP = D.
Self Assessment

2
1. In each of the following cases, let T be the linear operator on R which is represented by the
2
matrix A in the standard ordered basis for R , and let U be the linear operator on C 2
represented by A in the standard ordered basis. Find the characteristic polynomial for T

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