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I have been given a vector problem, np as I am good with vectors. But I was educated in Denmark, and I'm currently in America. The assignment is

Find $a^T\cdot b$.

Now I have never seen this $\{-\}^T$ before, what does it mean?

If it helps to explain, I have been given $a= [1,2,0]$ and $b = [2,0,4]$ and $4$ questions. Find $||a||$, Find $a^T\cdot b$, Find $a \times b$, Find $a\cdot b^T$

On a side note I, assume that $||a||$ is the length of a right?

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I think the $\cdot$ in "$\mathbf{a}^T \cdot \mathbf{b}$" is bad notation here and perhaps incorrect. $\cdot$ usually refers to the dot product. If $\mathbf{a}$ and $\mathbf{b}$ are column vectors of the same length, then $\mathbf{a}^T \mathbf{b}$ works out to be the dot product of $\mathbf{a}$ and $\mathbf{b}$ (strictly speaking, it is the 1-by-1 matrix containing whose element is that dot product). The $\cdot$ is not needed. In your problem, apparently $\mathbf{a}$ and $\mathbf{b}$ are row vectors, so $\mathbf{a}^T \mathbf{b}$ is a 3-by-3 matrix. –  Stefan Smith Nov 4 '13 at 0:38
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2 Answers

up vote 0 down vote accepted

They are considered as matrices, usually column vectors, and $^T$ means transpose, i.e. exchanging the rows and columns (exchanging the indices: $(a_{ij})^T:=(a_{ji})$.) So, $a^Tb$ is the scalar product $\langle a,b\rangle$, and $ab^T$ will be a (rank 1) matrix of size $3\times 3$.

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If $a$ and $b$ are row vectors, then you have the inner and outer products the wrong way around. –  Daryl Sep 30 '12 at 23:14
So the correct answer to a^Tb will be [2] right? –  DoomStone Sep 30 '12 at 23:32
@DoomStone Is the original $a$ equal to $[1,2,0]$? If so, then $ab^T$ is $2$. But if the original $a$ is $\begin{bmatrix}1\\2\\0\end{bmatrix}$ (as is more common with North American mathematicians) then $a^Tb$ is $2$. –  alex.jordan Oct 1 '12 at 0:35
you are right.. –  Berci Oct 1 '12 at 11:31
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It could be that the $T$ indicates the transpose. If both $a$ and $b$ are row vectors as you have written them, then you will have to take transposes in order to multiply them.

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