Page 376 - A Course in Linear Algebra with Applications

360 Chapter Nine: Advanced Topics

and hence

AX = 2X, AY = 2Y + X and AZ = 2Z.

It is now evident that {X, Y, Z} is a basis of C 3 consisting
of the two Jordan strings X,Y, and Z. Therefore the Jordan
form of A has two blocks and is

/ 2 1 1 0
0 2 1 0
N

V 0 0 1 2

As an application of Jordan form we establish an inter-
esting connection between a matrix and its transpose.

Theorem 9.4.7
Every square complex matrix is similar to its transpose.

Proof
Let A be a square matrix with complex entries, and write N
1
for the Jordan normal form of A. Thus S~ AS = N for some
invertible matrix S by 9.4.5. Now

T T
T T
1 T
T
N T = S A (S- ) = S A (S )-\
T
so N T is similar to A . It will be sufficient if we can prove that
N and N T are similar. The reason for this is the transitive
property of similarity: if P is similar to Q and Q is similar to
R, then P is similar to R.
Because of the block decomposition of N, it is enough
to prove that any Jordan block J is similar to its transpose.
But this can be seen directly. Indeed, if P is the permutation
matrix with a line of l's from top right to bottom left, then
T
X
matrix multiplication shows that P~ JP — J .