Suppose $A$ is a diagonal matrix. I want to show that $A$ is normal. That is: $A^* A = A A^* $. Since $A$ is diagonal, then $A = A^*$ and so $A$ is trivially normal. IS this correct? or I am missing something?
Asked
Active
Viewed 1,879 times
0
-
Your matrix is real , right ? – Nizar Oct 27 '15 at 10:41
-
not necesarilly – Oct 27 '15 at 10:42
-
1For example, note that $$ \pmatrix{i\&i}^* = \pmatrix{-i\&-i} $$ – Ben Grossmann Oct 27 '15 at 15:09
2 Answers
3
It is not true that $A=A^{\ast}$ is general; this is true if and only if $A$ is real.
In multiplying diagonal matrices all that happens is multiplying the diagonal elements pairwise. This is obviously commutative, so multiplication of diagonal matrices is commutative.
Servaes
- 63,261
- 7
- 75
- 163
-
1
-
It is true for any matrix that, $A$ is normal if and only if $A$ is diagonalizable right?? – Fareed Abi Farraj Jul 17 '19 at 04:59
-
1@FareedAF No; any normal matrix is diagonalizable, but not every diagonalizable matrix is normal. See this question for details. – Servaes Jul 17 '19 at 08:46
1
If $A$ is real, then you're right.
If $A$ is complex, then $A$ is not the same as $A^*$.
In any case, $A^* A = A A^*$ because $A^* A$ and $A A^*$ are diagonal matrices with entries $\overline a_{ii} a_{ii}=a_{ii} \overline a_{ii}$.
lhf
- 216,483