# diagonalizable

• Apr 15th 2011, 07:06 AM
alexandrabel90
diagonalizable
in my book, it states that the 2x2 matrix,

(1 2i
2 1) is unitarily diagonalizable. i thought it has to be normal to be unitarily diagonalizable? in this case, the matrix doesnt seem to be normal.

may i know what is wrong with my reasoning?
• Apr 15th 2011, 07:27 AM
Swlabr
Quote:

Originally Posted by alexandrabel90
in my book, it states that the 2x2 matrix,

(1 2i
2 1) is unitarily diagonalizable. i thought it has to be normal to be unitarily diagonalizable? in this case, the matrix doesnt seem to be normal.

may i know what is wrong with my reasoning?

It is normal...(you should remember that a matrix is normal if and only if it is unitarily diagonalisable).

It is normal as if

(1 2i)
(2 1)=A

$A=\left( \begin{array}{cc}
1 & 2i \\
2 & 1 \end{array} \right)$

and

(1 2)
(-2i 1)=B

$B=\left( \begin{array}{cc}
1 & 2 \\
-2i & 1 \end{array} \right)$

then A and B commute. As B is the conjugate transpose thing of A, you are done...

(I've left the LaTeX in in the hope that the compiler will start working again...)
• Apr 15th 2011, 07:31 AM
alexandrabel90
shouldnt
(1 2
-2i 1) =B?
• Apr 15th 2011, 07:34 AM
Swlabr
Quote:

Originally Posted by alexandrabel90
shouldnt
(1 2
-2i 1) =B?

Yes, sorry, the minus sign went walkabout between the LaTeX and the Ascii stuff. What did you get when you calculated AB and BA?
• Apr 15th 2011, 07:50 AM
alexandrabel90
i got that

(5 2+2i
2-2i 5) = AB

(5 2i+2
2i+2 5)=BA

thus dont commute but it should commute :(
• Apr 15th 2011, 07:53 AM
Swlabr
Quote:

Originally Posted by alexandrabel90
i got that

(5 2+2i
2-2i 5) = AB

(5 2i+2
2i+2 5)=BA

thus dont commute but it should commute :(

You BA is incorrect, but your AB is correct. Looks like a simple case of a wandering minus sign to me...