If I know rank-$1$ matrix ${\bf x}{\bf x}^\top$, how can I find $\| {\bf x} \|_2^2$?

Question

If $A$ is a $3 \times 1$ matrix and $$AA^{T} = \begin{pmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \\ \end{pmatrix}$$ what is $A^{T}A$?

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I have tried, transposing both sides of the equation, $(AA^{T})^{T}$ , and transposing the matrix as well, but it doesn't give a 1x1 matrix. I am unsure of which property to use! Any help will be greatly appreciated!

Answer 1

Try the following: Calculate $$ (AA^\top)^2 = AA^\top A A^\top. $$ Then observe that $A^\top A$ is a number and we can pull it to the front: $$ (A^\top A)AA^\top $$ So you just have to observe the common factor in $(AA^\top)^2$.

Answer 2

The trace has the property that $\mbox{tr}(AB) = \mbox{tr}(BA)$. Therefore, $\mbox{tr} \left( A^T A \right) = 3$, and then

$$A^T A = (3)$$