Page 162 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes T i c i i … (1)
for i = 1 to n then the matrix of the linear transformation is given by

C 1 0 0 0 0 ... 0
0 C 2 0 0 0 ... 0
D = 0 0 C 3 0 0 ... 0 ...(2)
      
0 0 0 ... ... ... C
n
with the help of equation (2) we would gain considerable information about T. Simple numbers
associated with T, such as the rank of T or the determinant of T, would be determined with little
more than a glance. The range of T would be the subspace spanned by those s for which c 0,
i i
and the null space would be spanned by the remaining s. Indeed, it seems fair to say that, if we
i
knew a basis and a diagonal matrix D such that [T] = D, we could answer readily any question
about T which might arise.
In the following we are interested in finding out if a linear operator can be represented by a
diagonal matrix. How can we find the basis for such type of linear operator and what are the
values of c s.
i
12.2 Characteristic Values

Guided by the equation (1) we should study vectors which on application of linear operator T
transformed into the scalar multiples of themselves.

Let V be a vector space over the field F and T be a linear operator on V. A characteristic value of
T is a scalar C in F such that there is a non-zero vector in V with T = c . If c is a characteristic
value of T, then
(a) Any such that T = c , is called characteristic vector of T.
(b) The collection of all such that T = c , is called the characteristic space associated with c.
If T is any linear operator and c is any scalar, the set of vectors , such that T = c is a sub-space
of V. It is null space of linear transformation (T– cI). We call c a characteristic value of T if this
subspace is different from the zero subspace, i.e., if (T – cI) fails to be 1:1. If the underlying space
V is finite-dimensional, (T – cI ) fails to be 1:1 precisely when its determinant is different from 0.
Theorem 1: Let T be a linear operator on a finite-dimensional space V and let c be a scalar. The
following are equivalent:
(i) c is a characteristic value of T.
(ii) The operator (T – cI) is singular (not invertible)

(iii) det (T – cI) = 0.
The determinant criterion (iii) is very important because it tells us where to look for the
characteristic values of T. Since det (T – cI) is a polynomial of degree n in the variable c, we will
find the characteristic values as the roots of that polynomial.
If  is any ordered basis of V and A= [T] , then (T – cI) is invertible if and only if the matrix
(A – cI) is invertible. Accordingly, we make the following definition.

Definition: If A is an n n matrix over the field F, a characteristic value of A in F is a scalar c in
F such that the matrix (A – cI) is singular (not invertible).

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