## Why is a Linear Transformation a Type-(1,1) Tensor?

I've seen in multiple sources that a linear transformation constitutes a tensor with one contravariant and one covariant index. Could someone explain to me why this is the case? I'm asking not because I have a solid understanding of tensors and am confused about this particular example; rather, I have a very solid understanding of linear transformations and am trying to use this knowledge to understand the co- and contravariance of indices. I understand the co- and contravariance of vectors, but something about applying this concept to indices has created a gap in my understanding.

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Mentor
 Quote by marschmellow a linear transformation constitutes a tensor with one contravariant and one covariant index.
It doesn't, but it can be used to define one. If L is the linear transformation (from V into V), then you can define a (1,1) tensor T by T(u*,v)=u*(L(v)) for all u* in V* and all v in V.

 Quote by Fredrik It doesn't, but it can be used to define one. If L is the linear transformation (from V into V), then you can define a (1,1) tensor T by T(u*,v)=u*(L(v)) for all u* in V* and all v in V.
Okay I can see that, but then why do people refer to a linear transformation as a (1,1) tensor? Now that I think about it, it doesn't seem like a tensor at all--it maps a vector to a vector, not some vectors to a scalar. The tensor you created maps a vector and a covector to a scalar, but you had to add something (a linear functional) to the linear transformation to get it.

Mentor

## Why is a Linear Transformation a Type-(1,1) Tensor?

 Quote by marschmellow Okay I can see that, but then why do people refer to a linear transformation as a (1,1) tensor?
Because the vector space of linear maps from V into V is isomorphic to the vector space of (1,1) tensors.

 Quote by Fredrik Because the vector space of linear maps from V into V is isomorphic to the vector space of (1,1) tensors.
Why just (1,1) tensors? Isn't the space of all linear maps V-->V isomorphic to any [dim(V)]^2- dimensional vector space? Do the vector spaces of (2,0) tensors and (0,2) tensors not automatically satisfy this condition?

 Mentor Yes, for all integers n, any two n-dimensional vector spaces are isomorphic. However, in the general case, you need something like a basis for each of the spaces to explicitly define the isomorphism. In this case, you just define, for each linear L:V→V, $$T_L(u^*,v)=u^*(L(v))$$ for all $u^*\in V^*$ and all $v\in V$, and the map $L\mapsto T_L$ is an isomorphism. No need to use any special bases, inner products, metrics or anything of that sort. Another reason is that this identification between those two spaces works so well with the abstract index notation. In that notation, both L and TL above would be denoted by $L^a{}_b$. The map I denoted by L is $v^b\mapsto L^a{}_b v^b$, and the map I denoted by TL is $(\omega_a,v^b)\mapsto L^a{}_b \omega_a v^b$. (Here I'm denoting a typical member of V* by a Greek letter instead of by a Latin letter with an asterisk, because the asterisks would make the abstract index notation awkward). I should probably tell you that I'm making another identification by isomorphism in the preceding paragraph. $L^a{}_bv^b$ has a free index upstairs, so it should be interpreted as a (1,0) tensor, i.e. a member of V**. Specifically, it's the map $u^*\mapsto T_L(u^*,v)$, which could be denoted by $T_L(\cdot,v)$. But I'm using the isomorphism $f:V\to V^{**}$ defined by $f(v)(u^*)=u^*(v)$ for all $v\in V$ and $u^*\in V^*$, to identify V with V**. Since $$f(L(v))(u^*)=u^*(L(v))=T_L(u^*,v)=T_L(\cdot,v)(u^*)$$ for all $u^*\in V^*$, we have $f(L(v))=T_L(\cdot,v)$. This means that the map $T_L(\cdot,v)$ is the member of V** that the isomorphism f associates with $L(v)\in V$. So when I said that L is denoted by $L^a{}_b$ in the abstract index notation, that was strictly speaking not true, but I said it anyway, because you will see lots of statements like that in GR books. It's more accurate to say that $L^a{}_b$ takes $v^b$ to $L^a{}_bv^b$, which denotes $T_L(\cdot,v)=f(L(v))$, where f is the natural isomorphism from V to V**. Yes, this is very confusing at first. It takes some time getting used to. I think the key to understanding these things is to understand the isomorphism I denoted by f, so if you don't understand it perfectly, you should start by making sure that you do.

 Quote by Fredrik Yes, for all integers n, any two n-dimensional vector spaces are isomorphic. However, in the general case, you need something like a basis for each of the spaces to explicitly define the isomorphism. In this case, you just define, for each linear L:V→V, $$T_L(u^*,v)=u^*(L(v))$$ for all $u^*\in V^*$ and all $v\in V$, and the map $L\mapsto T_L$ is an isomorphism. No need to use any special bases, inner products, metrics or anything of that sort. Another reason is that this identification between those two spaces works so well with the abstract index notation. In that notation, both L and TL above would be denoted by $L^a{}_b$. The map I denoted by L is $v^b\mapsto L^a{}_b v^b$, and the map I denoted by TL is $(\omega_a,v^b)\mapsto L^a{}_b \omega_a v^b$. (Here I'm denoting a typical member of V* by a Greek letter instead of by a Latin letter with an asterisk, because the asterisks would make the abstract index notation awkward). I should probably tell you that I'm making another identification by isomorphism in the preceding paragraph. $L^a{}_bv^b$ has a free index upstairs, so it should be interpreted as a (1,0) tensor, i.e. a member of V**. Specifically, it's the map $u^*\mapsto T_L(u^*,v)$, which could be denoted by $T_L(\cdot,v)$. But I'm using the isomorphism $f:V\to V^{**}$ defined by $f(v)(u^*)=u^*(v)$ for all $v\in V$ and $u^*\in V^*$, to identify V with V**. Since $$f(L(v))(u^*)=u^*(L(v))=T_L(u^*,v)=T_L(\cdot,v)(u^*)$$ for all $u^*\in V^*$, we have $f(L(v))=T_L(\cdot,v)$. This means that the map $T_L(\cdot,v)$ is the member of V** that the isomorphism f associates with $L(v)\in V$. So when I said that L is denoted by $L^a{}_b$ in the abstract index notation, that was strictly speaking not true, but I said it anyway, because you will see lots of statements like that in GR books. It's more accurate to say that $L^a{}_b$ takes $v^b$ to $L^a{}_bv^b$, which denotes $T_L(\cdot,v)=f(L(v))$, where f is the natural isomorphism from V to V**. Yes, this is very confusing at first. It takes some time getting used to. I think the key to understanding these things is to understand the isomorphism I denoted by f, so if you don't understand it perfectly, you should start by making sure that you do.
I appreciate this very in-depth reply. I'll definitely be taking a hard look at this soon. Thanks!