# How can I see this special case of tensor definition as a definition of vectors(quantities with both magnitude and direction)in $\mathbb R^3$?[Edited]

Doubt:-

What is the physical meaning of tensor? How can I see this special case of tensor definition as a definition of vectors(quantities with both magnitude and direction)in $\mathbb R^3$? Please help me. I could see $L_0(V)$ as a set of scalars. So, $k-\text{tensors on } V$ is a generalization of scalars. When $k=0$ tensors are scalars. How can I see vectors like this?

– user464147
Jul 5, 2018 at 2:54
• A $k$-tensor is essentially a matrix represented in $k$ dimensions, so a 'vector is a $1$-tensor and a matrix is a $2$- tensor. A confusing issue is that in physics people say tensor when they mean what a mathematician would call a tensor-valued function or a tensor field.
– Ian
Jul 5, 2018 at 3:26
• You mean 2-tensor means $Row 1-[f(e_1,e_1) f(e_1, e_2)], Row 2-[f(e_2,e_1) f(e_2, e_2)]$ like this?
– user464147
Jul 5, 2018 at 3:37
• How it is the generalisation of ordinary vectors?
– user464147
Jul 5, 2018 at 3:38
• the set of all tensors is also a vector space Jul 5, 2018 at 4:51

0-tensors are constant functions, which we identify with scalars.

1-tensors are linear functions, which we identify with vectors. This identification amounts to selecting an inner product: we identify the vector $x$ with the function $y \mapsto \langle x,y \rangle$.

2-tensors are bilinear functions, which we identify with matrices. This identification also amounts to selecting an inner product: we identify the matrix $A$ with the function $(x,y) \mapsto \langle x,Ay \rangle$.

Things become a bit foreign when we go to $k$-tensors with $k>2$. One way to think about it is that a $k$-tensor takes a vector and gives back a $(k-1)$-tensor. Thus for instance a $3$-tensor takes a vector and gives back a matrix.

• How do you guarantee that $V$ is Inner product space?
– user464147
Jul 5, 2018 at 15:47
• You don't, but selecting an inner product provides such an identification. Without selecting an inner product, tensors remain abstract linear functions.
– Ian
Jul 5, 2018 at 16:01
• Now i have one more doubt, stress tensors are actually bilinear. Right?en.wikipedia.org/wiki/Cauchy_stress_tensor
– user464147
Jul 5, 2018 at 16:21
• That is correct, though as I mentioned physicists often say tensor when they mean tensor field.
– Ian
Jul 5, 2018 at 16:48
• Thank you very much @Ian
– user464147
Jul 5, 2018 at 17:00