Page 314 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes
Then = E j and
j
T
f(T) = f ( ) E j
j
c
= f ( )E j
j
j
= bE j
j
Hence,
2
2 2
c
f
c
( ( ) b )E j = f ( ) b E j
j
j
j j
= 0.
Therefore, f(c ) = b or E = 0. By assumption, 0, so there exists an index i such that E 0. It
j j i
follows that f(c ) = b and hence that f(S) is the spectrum of f(T). Suppose, in fact, that
i
f(S) = {b , ... , b }
1 r
where b b when m n. Let X be the set of indices i such that 1 i k and f(c ) = b . Let P = E i
m n m i m m
j
the sum being extended over the indices i in X . Then P is the orthogonal projection of V on the
m m
subspace of characteristic vectors belonging to the characteristic value b of f(T), and
m
r
f(T) = b P
m m
m 1
is the spectral resolution of f(T).
–1
Now suppose U is a unitary transformation of V onto V' and that T’ = UTU . Then the equation
T = C
holds if and only if
T'U = cU .

Thus S is; the spectrum of T', and U maps each characteristic subspace for T onto the corresponding
subspace for T'. In fact, using (2), we see that

T' = c E , E j UE U 1
j j
j
j
is the spectral resolution of T'. Hence
c
f(T’) = f ( )E j
j
j
1
= f ( )UE U
c
j
j
j
1
= U f ( )E ) U
c
j j
j
= Uf(T) –1

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