I am trying to find out if the operation is valid.
Suppose we have a vector $z$ such that $Az$ is defined and valid. Is $z^TA$ also valid? Let A be square, for this case. It seems to create a row vector.
I am trying to find out if the operation is valid.
Suppose we have a vector $z$ such that $Az$ is defined and valid. Is $z^TA$ also valid? Let A be square, for this case. It seems to create a row vector.
Yet it is valid. Let $z\in\Bbb{R}^{n\times 1}$ and $A\in\Bbb{R}^{n\times n}$. Then $Az\in\Bbb{R}^{n\times 1}$ and $z^\top A\in\Bbb{R}^{1\times n}$. Actually, when you use vectors you need to state that your vector space contains column or row vectors. Personally, I prefer the first one. Then, you can just state that $z\in\Bbb{R}^n$ and you will mean that $z$ is an $n$-dimensional column (or row) vector.