# Confusion with tensors

1. Apr 8, 2015

### JonnyG

First let me give the definition of tensor that my book gives:

If $V$ is a finite dimensional vector space with $dim(V) = n$ then let $V^{k}$ denote the k-fold product. We define a k-tensor as a map $T: V^{k} \longrightarrow \mathbb{R}$ such that $T$is multilinear, i.e. linear in each variable if all the other variables are held fixed. I know there are more general definitions but since this is the one I am using in my book, let's stick with this one.

Okay, now here is my problem. First off, assume from this point on that $\mathbb{R}^n$ has the usual basis. If $L^{2}(\mathbb{R}^{n})$ is the set of all 2-tensors on $\mathbb{R}^n$ then it has a dimension of $n^2$. If $M(n,n)$ is the set of all n by n matrices with real entries then we have $L^2(\mathbb{R}^n) \cong M(n,n)$.

However, let $L(\mathbb{R}^n, \mathbb{R}^n)$ be the set of all linear transformations $f: \mathbb{R}^n \longrightarrow \mathbb{R}^n$. It seems obvious to me that $L(\mathbb{R}^n, \mathbb{R}^n) \cong M(n,n)$. But then this would imply that $L^2(\mathbb{R}^n) \cong L(\mathbb{R^n}, \mathbb{R}^n)$ which is impossible since $dim(L^2(\mathbb{R}^n)) = n^2$ and $dim(L(\mathbb{R}^n, \mathbb{R}^n)) = n$ (I proved its dimension is $n$ and I am sure the proof is correct).

Where have I gone wrong?

2. Apr 8, 2015

### Fredrik

Staff Emeritus
The correspondence between $L(\mathbb R^n,\mathbb R^n)$ and $M(n,n)$ is bijective. So these vector spaces must have the same dimension.

3. Apr 8, 2015

### JonnyG

That's what I thought as well. So this must just mean my "proof" that $dim(L(\mathbb{R}^n, \mathbb{R}^n)) = n$ is incorrect.

4. Apr 8, 2015

### Fredrik

Staff Emeritus
Yes, I agree.

5. Apr 8, 2015

### WWGD

I think you need to specify the target space of $L^2 (\mathbb R^n, \mathbb R^n)$ , i.e., bilinear maps into what space? If it is into the base field $\mathbb R$ of dimension 1 as a v.space over itself, then the dimension is $n^2$ as you said, if the target space is another vector space V, then $Dim L^2 (\mathbb R^n , \mathbb R^n, V )$ is the product of the individual dimensions. Sorry if this is obvious and I misread what you meant.

6. Apr 8, 2015

### Fredrik

Staff Emeritus
He never mentioned $L^2(\mathbb R^n,\mathbb R^n)$. The question was about $L(\mathbb R^n,\mathbb R^n)$ (linear operators on $\mathbb R^n$) and $L^2(\mathbb R^n)$ (bilinear maps from $\mathbb R^n\times\mathbb R^n$ into $\mathbb R$).

7. Apr 8, 2015

### WWGD

OK. Is $L^2(\mathbb R^n)$ supposed to be maps into $\mathbb R$, or into any vector space ( as in the case of matrix multiplication)?

8. Apr 8, 2015

### JonnyG

$L^2(\mathbb{R}^n)$ maps into the reals. I see the mistake in my "proof" that led to this confusion. It all makes sense now.

9. Apr 8, 2015

### WWGD

Just curious, what was it?

10. Apr 8, 2015

### JonnyG

***Sorry about the crappy LaTex ***

Well the correct proof would be to fix the usual basis for $\mathbb{R}^n$. Let $T \in L(\mathbb{R}^n,\mathbb{R}^n)$. Let $\phi_{j,i} (e_i) = e_j$ but be 0 on all other basis vectors. Note that $1 \le j \le n, 1 \le i \le n$. We show that the $\phi_{j,i}$ span $L(\mathbb{R}^n,\mathbb{R}^n)$. We know that $T(e_j) = \sum_{j = 1}^n d_{j,i} e_j$ for some scalars $d_{j,i} \in \mathbb{R}$.

So $T(e_1) = d_{1,1} \phi_{1,1} (e_1) + d_{2,1} \phi_{2,1} (e_1) + d_{3,1} \phi_{3,1} (e_1) + \cdots + d_{n,1} \phi_{n,i} (e_n)$

We do this for each $e_i$. Clearly then, $T$ is a linear combination of the $\phi_{j,i}$. It is also clear that there are $n^2$ many of the $\phi_{j,i}$ and Linear independence is obvious.

The mistake I was making before was only allowing one index, so I ended up with only $n$ many of the $\phi$

11. Apr 8, 2015