Thread: triangularize the operator

+riangularize the operator T: C^3 --> C^3 given by T(x,y,z)= (x, yi, y+zi). Find the ordered basis B and the upper triangular matrix Mat(T, B ,B).

This is what I have so far, please let me know how it looks. Thanks!


Standard basis : (1,0,0)=(1,0,0)
(0,1,0)=(0, i, 1)
(0,0,1)=(0,0, i)
Apply i times the second row and add to the third.
Now I have (1,0,0)
(0, i, 0)
(0,1, i)
So the eigen values are {1,i}

Originally Posted by jax

+riangularize the operator T: C^3 --> C^3 given by T(x,y,z)= (x, yi, y+zi). Find the ordered basis B and the upper triangular matrix Mat(T, B ,B).

This is what I have so far, please let me know how it looks. Thanks!


Standard basis : (1,0,0)=(1,0,0)
(0,1,0)=(0, i, 1)
(0,0,1)=(0,0, i)


What do you mean with these weird equalities?? The 2nd and 3rd. ones are obviously false...

Take simply the standard basis ...

Apply i times the second row and add to the third.
Now I have (1,0,0)
(0, i, 0)
(0,1, i)
So the eigen values are {1,i}

The matrix representing T wrt the standard basis above is just , and from it is immediate that the eigenvalues

of the matrix are, indeed, ,as you said. Now, the matrix is diagonalizable iff the eigenspace belonging to has dimension 2...

Tonio