# Thread: triangularize the operator

1. ## 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}

2. 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 $\displaystyle \left\{(1,0,0), (0,1,0), (0,0,1)\right\}$...

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 $\displaystyle \begin{pmatrix}1&0&0\\0&i&0\\0&1&i\end{pmatrix}$ , and from it is immediate that the eigenvalues

of the matrix are, indeed, $\displaystyle 1,i$ ,as you said. Now, the matrix is diagonalizable iff the eigenspace belonging to $\displaystyle i$ has dimension 2...

Tonio