# How do you prove the rank of a second order tensor is equivalent to the rank of the corresponding matrix.

I hear this said in a lot of places but I can't find a proof of it anywhere. Let $V$ and $W$ be vector spaces of dimension $m$ and $n$ respectively. Then for any $t \in V \otimes W$, we have

$$t = v_1 \otimes w_1 + \cdots v_k \otimes w_k$$

The rank of $t$ is defined to be the smallest number $r$ such that $t$ can be written as $r$ simple tensors, ie. tensors of the form $v\otimes w$.

The tensor $t$ apparently corresponds to the matrix that is the sum of the outer products. $$v_1 w_1^T + \cdots v_kw_k^T$$ and the definition of the rank of $t$ coincides with the rank of this matrix.

Can someone show me a proof that these two notions of rank coincides or show me where I can see a proof?

This only really makes sense if $V = F^m$ and $W = F^n$ where $F = \mathbb R, \mathbb C, \mathbb Q$, or whatever is the field you do your linear algebra over, so I'm just gonna assume that $v_i, w_i$ are $m \times 1$ and $n \times 1$ column vectors.

Note that matrix multiplication is bilinear so the map $\phi\colon F^m \otimes F^n \to \mathbb M_{m,n}(F)$ that sends a simple tensor $v \otimes w$ to $vw^T$ is well defined. In fact this is an isomorphism: The standard basis of $F^m \otimes F^n$ is to take all $e_i \otimes e_j$ where $e_i$ has a $1$ in the $i^\text{th}$ row and zeros elsewhere and $e_i \otimes e_j$ gets sent to $E_{ij}$, the matrix with a $1$ in the $(i, j)^\text{th}$ position and zero's elsewhere. The $E_{ij}$ are a basis of $\mathbb M_{m,n}(F)$ so $\phi$ is an isomorphism.

Now if $t \in F^m \otimes F^n$ has rank $k$ then it can be written as $$t = v_1 \otimes w_1 + \cdots + v_k \otimes w_k$$ and the matrix $\phi(t)$ has rank at most $k$ because every column is a linear combination of the $v_i$.

On the other hand if the matrix $\phi(t)$ has rank $k$ then let $v_1, \ldots, v_k$ be a basis for it's column space. Then it's not hard to see that there exist vectors $w_1, \ldots, w_k$ such that $\phi(t) = v_1w_1^T + \cdots + v_kw_k^T$ ($w_i$ is just the coefficients of $v_i$ when the columns of $\phi(t)$ are written as linear combinations of the $v_i$). As $\phi$ is an isomorphism $t = v_1 \otimes w_1 + \cdots + v_k \otimes w_k$ which means the rank of $t$ is at most $k$.

Each definition of rank is at most the other, so they're equal.

• Thanks this is pretty helpful. Can you just tell me why this only makes sense if $V$ and $W$ have the same dimension. Couldn't you define the map $\phi: \mathbb{F}^m \otimes \mathbb{F}^n \rightarrow \mathbb{M}_{m \times n}$ and proceed the same way. I think this map should be an isomorphism since these are both vector spaces of dimension $mn$. – jlc1112 Mar 13 '15 at 16:16
• I think I was just confused when I wrote that, the dimensions don't need to be the same. I'll edit. – Jim Mar 13 '15 at 21:26