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

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 M_{2}(R) can be naturally identified with R^{4}.

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:M_{2}(R) --> M_{2}(R) is given by the following formula:

T(a,b,c,d) = (a,c,b,d).

Hence T is a linear map of R^{4}=M_{2}(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) = A^{t}

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

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:M

T(a,b,c,d) = (a,c,b,d).

Hence T is a linear map of R

T = 1 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

That is T(A) = A

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)

| 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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