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Answer to Question #4129 in Linear Algebra for Myungjin Choi

Question #4129
If T is a transpose map, such that T(A) = transpose of A for
any 2 by 2 matrix A with real entries, find the eigenvalues of T and a basis of M_2(R) (where M_2(R) denote all the 2 by 2 matrices with real entries) with respect to which T is diagonal
Expert's answer
The space M2(R) can be naturally identified with R4.
Thus every real 2x2 matrix
A= a b
c d
can be written as a vector with coordinates A = (a,b,c,d).

Then the transpose map T:M2(R) --> M2(R) is given by the following formula:
T(a,b,c,d) = (a,c,b,d).
Hence T is a linear map of R4=M2(R) given by the following matrix:
T = 1 0 0 0
0 0 1 0
0 1 0 0
0 0 0 1
That is T(A) = At
It is easy to see that the characteristic polynomial of T is
| 1-t 0 0 0 |
| 0 -t 1 0 |
| 0 1 -t 0 | = (t-1)3(t+1)
| 0 0 0 1-t |

So T has two eignevalues:
t=1 and t=-1
Let us find eigenvectors of T.
For t=-1 we haveto solve the equation
(T+E)v =
| 2 0 0 0 |
| 0 1 1 0 | = 0
| 0 1 1 0 |
| 0 0 0 2 |

we find eigen vector
v=(0,1,-1,0)
corresponding to the matrix
v =
0 -1
1 0
For t=1, we have to solve the equation
(T-E)v =
| 0 0 0 0 |
| 0 -1 1 0 |
| 0 1 -1 0 |
| 0 0 0 0 |
= 0
which gives three linearly independent vectors:
w1 = (1,0,0,0)

w2 = (0,1,1,0)
w3 = (0,0,0,1)
corresponding to matrices
w1 =
1 0
0 0
w2 =
0 1
1 0
w3 =
0 0
0 1
Then the matrices
w1, w2, w3, v
constitute the basis in which T has the following diagonal form:
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 -1

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