Page 167 - DMTH502_LINEAR_ALGEBRA

Unit 12: Introduction and Characteristic Values of Elementary Canonical Forms

Solution: The characteristic equation of the square matrix A is given by Notes
|A – I| = 0
Similarly the characteristic equation of the matrix
A’ is (A’ – I| = 0

Now, we have to prove that the characteristic roots of |A – I| = 0 and |A’ – I| = 0 identical.
Since interchange of row and column does not change the value of the determinant, hence we
have
|A – I| = |A’ – I|
Hence the roots of the equations |A – I| = 0 and |A’ – I| = 0 are same.

Lemma: If F is a characteristic value of T, then for any polynomial q(x) F(x), q( ) is a
characteristic value of q(T).
Proof: Suppose F and T = for non-zero vector in V. Now T = T(T ) = T( ) = T =
2
2 , continuing in this way we obtain T = 3 , T , T = 4 , ... T = k , for all positive
4
k
3
4
integers k. If
q(x) = a x = a x m–1 + ... + a F, then
m
0 1 m
m
q(T) = a T = a T m–1 + ... + a ,
0 1 m
hence q(T) = a m + a m–1 + ... + a
0 1 m
= q( )d.
Thus [q(T) – q( )I] = 0, since 0 so q( ) is characteristic value of q(T).
Definition: Let T be a linear operator on the finite dimensional space V. We say that T is
diagonalizable if there is a basis for V each vector of which is a characteristic vector of T.
The reason for the name should be apparent; for, if there is an ordered basis  = { , ...,  } for V
1 n
in which each  is a characteristic vector of T, then the matrix of T in the ordered basis  is
i
diagonal. If T = c , then
i i i
c 1 0 ... 0
0 c 2 ... 0
[ ]    
T
0 0 ... c n
We certainly do not require that the scalars c , ... c be distinct; indeed, they may all be the same
1 n
scalar (when T is a scalar multiple of the identity operator).
One could also define T to be diagonalizable when the characteristic vectors of T span V. This is
only superficially different from our definition, since we can select a basis out of any spanning
set of vectors.
n
For Examples 1 and 2 we purposely chose linear operators T on R which are not diagonalizable.
2
In Example 1, we have a linear operator on R which is not diagonalizable, because it has no
characteristic values. In Example 2, the operator T has characteristic values; in fact, the characteristic
polynomial for T factors completely over the real number field: f = (x – 1) (x – 2) . Nevertheless
2
T fails to be diagonalizable. There is only a one-dimensional space of characteristic vectors
associated with each of the two characteristic values of T. Hence, we cannot possibly form a basis
3
for R which consists of characteristic vectors of T.
Suppose that T is a diagonalizable linear operator. Let c , ... c be the distinct characteristic values
1 k
of T. Then there is an ordered basis  in which T is represented by a diagonal matrix which has

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