# $(1,1)$ tensor vs a linear transformation (matrix)

Take $d$-dimensional Vector space $V$ with Field $R$.

A typical linear algebra linear transformation $V \to V$ can be represented by a $d \times d$ matrix $A$ such that for some $v,w \in V$, $Av=w$.

I'm learning about tensors, and I understand that a $(1,1)$ tensor $T$ is a linear transformation $V^* \times V \to R$. I've read that such a $(1,1)$ tensor is equivalent to such a matrix.

However, I find it very difficult to imagine what $V^*$ (the dual space, i.e. set of all maps $V\to R$) has to do with a simple linear transformation from $R^d$ to $R^d$.

Moreover, the tensor components apparently are defined as $T^i_{\space \space j}=T(\epsilon_i, e^j)$, where $e^j, \epsilon _i$ are the $d$ bases of $V$ and $V^*$ respectively. This means that if we would write $T$ as a 2-dimensional array, it would have nothing to do with a matrix as in linear algebra.

So how are these two concepts connected?

This post is related to my question, but it doesn't really go into the difference between the matrix and tensor form.

Given a linear map $$\alpha:V\to V$$ we can construct a bilinear form $$\tau:V^*\times V\to R$$, by taking $$\tau(f,v)=f(\alpha v)$$. (Note that $$f(\alpha v)$$ makes sense because $$v\in V$$ and $$\alpha:V\to V$$ so $$\alpha v\in V$$, and then $$f\in V^*$$ means $$f:V\to R$$, so $$f(\alpha v)\in R$$.)

Similarly given a bilinear form $$\tau':V^*\times V\to R$$ we can construct a map $$\alpha': V\to V$$ by noting that if $$v\in V$$ then $$\tau'(-,v):V^*\to R$$, and hence $$\tau'(-,v)\in V^{**}$$. Since $$V$$ is finite dimensional we have $$V^{**}\cong V$$ and hence we can define $$\alpha'(v)$$ to be the element of $$V$$ corresponding to $$\tau'(-,v)$$ in $$V^{**}$$. This means that $$f(\alpha' v)=\tau'(f,v)$$.

Hence given a map $$V\to V$$ we get a map $$V^*\times V\to R$$ and given a map $$V^*\times V\to R$$ we get a map $$V\to V$$ (and furthermore if we translate back and forth we end up where we started). So we can view linear maps $$V\to V$$ as "the same as" bilinear maps $$V^*\times V\to R$$.

Finally lets check the matrices are the same. Given a map $$\alpha:V\to V$$ its matrix is defined by $$A^i_{\;j}=\epsilon_j(\alpha e^i)$$, and given a map $$\tau:V^*\times V\to R$$ its matrix is defined by $$T^i_{\;j}=\tau(\epsilon_j,e^i)$$. So if we have $$\tau(f,v)=f(\alpha v)$$ then $$T^i_{\;j}=\tau(\epsilon_j,e^i)=\epsilon_j(\alpha e^i)=A^i_{\;j}$$.

• Maybe: "we can view linear maps $V \to V$ as 'the same as' bilinear maps $V^* \otimes V \to R$"? I think this approach risks being a little confusing, because one is given an element of $V^* \otimes V$, not directly a bilinear map $V^* \otimes V \to R$. Of course contraction allows one to deduce a bilinear map from such an element, but it's at least as much of an argument as you've given here to show that one can go the other way. Why not argue directly to identify elements of $V^* \otimes V$ with (finite-rank) maps $V \to V$? Commented Nov 15, 2020 at 20:03
• Thanks for the first point. As for the second, the OP's teacher has defined $V\otimes V^*$ to consist of the bilinear maps $V^*\times V\to\mathbb R$. Commented Nov 15, 2020 at 20:32
• Whoops, right you are. I obviously didn't read the question carefully enough. :-) Commented Nov 15, 2020 at 22:45

Take $v\in V$ and $w_*\in V^*$. You have a natural bijection $i$ between $V$ and $V^*$ (because the dimension is finite), hence you can obtain the corresponding $i(w_*)=w\in V$.

Consider a map $Q:V\times V\to \Bbb R$ given by $$Q(x,y) = T(x,i(y)).$$ Obviously, $Q$ is a bilinear map, hence it can be given by a matrix $M$: $$Q(x,y) = x^TMy.$$ We will call this matrix - by extension - the matrix of $T$.

• Ok. good, so now I know how to make a $V\times V \to R$ version of a given $V\to V$ transformation. How do I do it the other way around? i.e. we have a certain tensor $T$, s.t. $T(v,w) = \sum \sum v^i w_j T(e_i, e^j)$. This is probably embarrassingly simple, but I don't seem to know how to "remove" the "dot product" that's in that equation, and just turn it into a map from $v$ to $w$. Commented Feb 8, 2017 at 7:21
• If you have a bilinear form $T(v,w)$ over $V\times V$, you can obtain a corresponding linear operator $S:V\to V$. Take a map $v\to T(v,\cdot)$. The object $T(v,\cdot)$ is a linear functional over $V$, hance it is - up to some isomorphism - an element of $V$ itself. Therefore, the linear map $v\to T(v,\cdot)$ can be seen as $V\to V$, therefore, a matrix. Commented Feb 8, 2017 at 9:11
• If you consider $T(v,w)$ you will not get a map from $v$ to $w$, it is by considering the action of $T$ on these vectors that you obtain an associated linear operator. Commented Feb 8, 2017 at 9:12
• I don't agree with this. Even when $V$ is finite dimensional there's not a natural isomorphism with its dual. And the answer doesn't need one. Commented Feb 8, 2017 at 11:55
• @OscarCunningham Thanks! The author explicitly stated that $V$ was a linear space over $\Bbb R$, so we can safely equip $V$ with an inner product. The question of "canonicity" is, after all, a matter of taste - it is sufficient that this isomorphism exists (like in $V^{**}\cong V$ for finite-dimensioned case or $V'\cong V$ for Hilbert spaces). Commented Feb 8, 2017 at 14:24