Page 151 - Linear Algebra Done Right

Normal Operators on Real Inner-Product Spaces
is the transpose of the matrix above. Use
Of course, the matrix of T
∗
and T T (do it
matrix multiplication to compute the matrices of TT
∗
∗
now). Because T is normal, these two matrices must be equal. Equating 139
the entries in the upper-right corner of the two matrices you computed,
you will discover that bd = ab. Now b = 0 because otherwise T would
be self-adjoint, as can be seen from the matrix in 7.17. Thus d = a,
completing the proof that (a) implies (b).
Now suppose that (b) holds. We want to prove that (c) holds. Choose
any orthonormal basis (e 1 ,e 2 ) of V. We know that the matrix of T with
respect to this basis has the form given by (b), with b = 0. If b> 0,
then (c) holds and we have proved that (b) implies (c). If b< 0, then,
as you should verify, the matrix of T with respect to the orthonormal

basis (e 1 , −e 2 ) equals ab , where −b> 0; thus in this case we also
−ba
see that (b) implies (c).
Now suppose that (c) holds, so that the matrix of T with respect to
some orthonormal basis has the form given in (c) with b> 0. Clearly
the matrix of T is not equal to its transpose (because b = 0), and hence
T is not self-adjoint. Now use matrix multiplication to verify that the
matrices of TT ∗ and T T are equal. We conclude that TT ∗ = T T, and
∗
∗
hence T is normal. Thus (c) implies (a), completing the proof.
As an example of the notation we will use to write a matrix as a
matrix of smaller matrices, consider the matrix
 
1 1 2 2 2
 
 1 1 2 2 2 
 
 0 0 3 3 3  .
D =
 
 0 0 3 3 3 
 
0 0 3 3 3
We can write this matrix in the form Often we can
understand a matrix
A B
D = , better by thinking of it
0 C
as composed of smaller
where matrices. We will use
this technique in the
 
3 3 3
1 1 2 2 2 next proposition and in


A = , B = , C =  3 3 3  ,
1 1 2 2 2 later chapters.
3 3 3
and 0 denotes the 3-by-2 matrix consisting of all 0’s.