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

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.

• $(AA^T)^T=(A^T)^TA^T = AA^T$ – hra Jun 21 '17 at 15:46
• Plug $B = A$ into your first equation – Daniel Xiang Jun 21 '17 at 15:49

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.