A is $n\times n$ matrix over complex numbers. Does $AA^T = A^TA$ imply that A is normal? If not what will be a counterexample?
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Since one definition of a normal matrix is one that satisfies $AA^\ast=A^\ast A$, you just need a matrix where $AA^T=A^TA$, but $AA^\ast\not=A^\ast A$. – Michael Burr May 24 '16 at 11:52
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yes this is my definition of normal matrix – Sushil May 24 '16 at 11:55
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No this isn't true.
Recall the definition:
Let $A^* = \bar{A}^\mathrm{T}$ denote the conjugate matrix of $A$. Then $A \in \mathbb{C}^{n,n}$ is called normal if $AA^* = A^*A$.
We can easily see that the matrix (taken from here) $A = \pmatrix{ 2&i\\i&-2}$ is orthogonal because
$$AA^\mathrm{T} = \pmatrix{ 3&0\\0&3} =A^\mathrm{T}A,$$
but
$$AA^* = \pmatrix{ 5&-4i\\4i&5} \neq \pmatrix{ 5&4i\\-4i&5} = A^*A.$$
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I just asked whether it will be normal. But I need to ask whether it will be similar to a normal matrix – Sushil May 24 '16 at 13:54
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4It would be better to ask a new question and leave this one in its original form than to change the current question. Therefore, the question and answer pair will make sense. – Michael Burr May 24 '16 at 14:06