Since $(AB^T) ^T = BA^T, (AA^T)^T=A^T A$?

where A, B are matrices.

I am wondering if the second equation holds.

  • 2
    $(AA^T)^T=(A^T)^TA^T = AA^T$ – rapidracim Jun 21 '17 at 15:46
  • 1
    Plug $B = A$ into your first equation – Daniel Xiang Jun 21 '17 at 15:49
up vote 1 down vote accepted

No, it doesn't hold.

In more detail, $$(AB^T)^T = (B^T)^T A^T = B A^T$$

Now, if you substitute $A$ for $B$ in the equation, you get $A A^T$ as the final result.

no it doesn't

let $A = [0\;1]$

then $(AA^T)^T$ is $1$ but $A^TA$ is a 2 by 2 matrix

Suppose $A$ is a $n \times m$ matrix:

Then $A^TA$ is $m \times m$ and $AA^T$ is $n \times n$

and there is no way $(AA^T)^T = A^TA$

Maybe it is possible with square matrices? I suggest you try it out on a couple of $2\times 2$ matrices and see what you get.

And algebraically.

$(AB)^T = B^TA^T\\(AA^T)^T = (A^T)^TA^T = AA^T$

  • how does$(A^T)^TA^T=AA^T$? – Jacob Claassen Jun 21 '17 at 15:55
  • Don't you agree that $(A^T)^T = A$? – Doug M Jun 21 '17 at 15:57
  • I don't see it, what's the reasoning? – Jacob Claassen Jun 21 '17 at 16:02
  • The transpose is a reflection across the main diagonal of the matrix. If we reflect again we get the original matrix back. en.wikipedia.org/wiki/Transpose – Doug M Jun 21 '17 at 16:08

Note that $$ [AA^T]^T = [A^T]^T[A^T] = AA^T $$ If $A$ is not square, then $A$ and $A^T$ can have different sizes. Even if $A$ is square, verify that with $$ A = \pmatrix{0&1\\0&0} $$ we have $AA^T \neq A^TA$. In fact, any matrix satisfying $AA^T = A^TA$ is called a normal matrix.

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