Many people say that they are the same, even I can't find much difference in them except that a vector/ matrix can be multiplied by any scalar, but to multiply it with a vector in $\Bbb{R^1}$ the vector or matrix should be of the order $1\times n$. What's and why is there a difference in this case?
Difference between a vector in $\Bbb{R^1}$ and a scalar
linear algebramultivariable-calculusvector-spacesvectors
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A scalar is a number, like 3 or 0.227. It has a bigness (3 is bigger than 0.227) but not a direction. Or not much of one; negative numbers go in the opposite direction from positive numbers, but that's all. Numbers don't go north or east or northeast. There is no such thing as a north 3 or an east 3.
A vector is a special kind of complicated number that has a bigness and a direction. A vector like $(1,0)$ has bigness 1 and points east. The vector $(0,1)$ has the same bigness but points north. The vector $(0,2)$ also points north, but is twice as big as $(0,2)$. The vector $(1,1)$ points northeast, and has a bigness of $\sqrt2$, so it's bigger than $(0,1)$ but smaller than $(0,2)$.
For directions in three dimensions, we have vectors with three components. $(1,0,0)$ points east. $(0,1,0)$ points north. $(0,0,1)$ points straight up.
A scalar field means we take some space, say a plane, and measure some scalar value at each point. Say we have a big flat pan of shallow water sitting on the stove. If the water is shallow enough we can pretend that it is two-dimensional. Each point in the water has a temperature; the water over the stove flame is hotter than the water at the edges. But temperatures have no direction. There's no such thing as a north or an east temperature. The temperature is a scalar field: for each point in the water there is a temperature, which is a scalar, which says how hot the water is at that point.
A vector field means we take some space, say a plane, and measure some vector value at each point. Take the pan of water off the stove and give it a stir. Some of the water is moving fast, some slow, but this does not tell the whole story, because some of the water is moving north, some is moving east, some is moving northeast or other directions. Movement north and movement west could have the same speed, but the movement is not the same, because it is in different directions. To understand the water flow you need to know the speed at each point, but also the direction that the water at that point is moving. Speed in a direction is called a "velocity", and the velocity of the swirling water at each point is an example of a vector field.
I think the only other thing to know is that in one dimension, say if you had water on a long narrow pipe instead of a flat dish, vectors and scalars are the same thing, because in one dimension there is only one way to go, forwards. Or you can go backwards, which is just like going forwards a negative amount. But there is no north or east or northeast. So one-dimensional vectors are interchangeable with scalars: all the vector stuff works for scalars, if you pretend that the scalars are one-dimensional vectors.
If this isn't clear please leave a comment.
Sometimes, we just say that a $1\times 1$ matrix is the same as a scalar. Afterall, when it comes to addition and multiplication of $1\times 1$ matrices vs addition and multiplication of scalars, the only difference between something like $\begin{bmatrix}3\end{bmatrix}$ and $3$ is some brackets. Consider $$(3+5)\cdot 4 = 32 \\ (\begin{bmatrix} 3\end{bmatrix} + \begin{bmatrix} 5\end{bmatrix})\begin{bmatrix} 4\end{bmatrix} = \begin{bmatrix} 32\end{bmatrix}$$ The algebra works out exactly the same. So sometimes it's not ridiculous to think of $1\times 1$ matrices as just another way of writing scalars.
But if you do want to distinguish the two, then just think of the formula $a\cdot b = a^Tb$ as a way of finding out which scalar you get from the dot product of $a$ and $b$ and not literally the dot product value itself (which should be scalar). That is, we calculate the dot product of $\begin{bmatrix} 1 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 3 \\ 4\end{bmatrix}$ by using the formula $$\begin{bmatrix} 1 \\ 2\end{bmatrix}^T\begin{bmatrix} 3 \\ 4\end{bmatrix} = \begin{bmatrix} 1 & 2\end{bmatrix}\begin{bmatrix} 3 \\ 4\end{bmatrix} = \begin{bmatrix} 11\end{bmatrix}$$ and then say that this tells us that the dot product is really $11$. So the formula $a^Tb$ is just an algorithm we use to find the correct scalar.
You can view it either way. It doesn't really make a difference.
Best Answer
Referring to formal definitions, the main difference is that: