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Suppose $A$ is an $n\times n$ matrix, and wondering when $A A^T$ equals $A^T A$? Only when $A$ is symmetric?

regards, Lin

Michael Hardy
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Lin Ma
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    http://math.stackexchange.com/questions/158219/is-a-matrix-multiplied-with-its-transpose-something-special – WaveX Feb 03 '17 at 03:03
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    Perhaps this may help answer your question? – WaveX Feb 03 '17 at 03:04
  • @WaveX, nice reference and vote up! I think as long as A is a normal matrix, it is fine, not necessarily to be orthogonal matrix, correct? – Lin Ma Feb 03 '17 at 06:14
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    Honestly I'm not sure if being orthogonal is necessary or not. I've just started studying Linear Algebra this year. – WaveX Feb 03 '17 at 13:51
  • @WaveX, your answer looks like a professor. Nice work and looking forward to talking more with you on math exchange. :) – Lin Ma Feb 03 '17 at 20:12

3 Answers3

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Matrices with that property are called normal matrices.

See Normal Matrix

blancmange
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Symmetric matrices are not the only matrices that satisfy this property. There is a definition for the matrix that you describe:

If $A$ is a complex matrix that satisfies $A^{\ast}A=AA^{\ast}$, then we say A is a normal matrix. Note that $A^{\ast}$ represents $A$ adjoint, i.e. the complex conjugate transpose of $A$.

So if $A$ is just a real matrix and $A$ satisfies $A^tA=AA^t$, then $A$ is a normal matrix, as the complex conjugate transpose of a real matrix is just the transpose of that matrix.

For example, a $2\times 2$ counterclockwise rotation matrix by an angle $\theta$ is given by $B=\begin{bmatrix} \cos\theta&-\sin\theta \\ \sin\theta& \cos\theta \end{bmatrix}$. Notice $B$ is not a symmetric matrix, but does satisfy $B^tB=BB^t$.

Michael Hardy
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TASPlasma
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  • Thanks TASPlasma, vote up, could you elaborate a bit more what do you mean adjoint and complex conjugate? Do you mean the elements of matrix is complex number, not real number? – Lin Ma Feb 03 '17 at 06:11
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    Yes I can. Often times, we consider matrices with complex numbers as entries instead of just purely real numbers. If $a+bi$ is a complex number, then the complex conjugate of $a+bi$ is $a-bi$. The adjoint of a matrix with complex entries, transposes the matrix, and also replaces every complex entry with their respective complex conjugate. Note that if our matrix is strictly a real matrix (in other words only has real entries), then the adjoint of that matrix is exactly equal to the transpose of that matrix. – TASPlasma Feb 03 '17 at 07:40
  • Thanks for the nice reply TASPlasma, mark your reply as answer. – Lin Ma Feb 03 '17 at 20:12
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Suppose $A \in M_{n,n}(\mathbb{R})$ then

$$AA^T = A^TA \iff A \textrm{ is normal}$$

Maciej Caputa
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    Don't you mean $A$ is a normal matrix? An orthogonal matrix A is one that satisfies $A^tA=AA^t=I$, where $I$ is the identity matrix. Not all normal matrices are orthogonal matrices. – TASPlasma Feb 03 '17 at 03:08
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    Not necessarily. $A$ is normal in this case, but it need not be orthogonal (the implication only goes from right to left). – Dave Feb 03 '17 at 03:08
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    You're right. I will correct it. – Maciej Caputa Feb 03 '17 at 03:10
  • @MaciejCaputa, nice reply and vote up. But how orthogonal related here? I think as long as A is a normal matrix, it is fine, not necessarily to be orthogonal matrix, correct? – Lin Ma Feb 03 '17 at 06:13
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    Yes, it is enough for $A$ to be normal matrix to satisfy the equation.

    Conversely, every matrix satisfying this equation must be normal but doesn't have to be orthogonal

     – Maciej Caputa Feb 03 '17 at 17:33
  • Thanks for the reply Maciej, vote up. – Lin Ma Feb 03 '17 at 20:12