I have to prove an equality between matrices $R=OTDO$ where

  • $R$ is a $M\times M$ matrix
  • $O$ is a $2\times M$ matrix
  • $T$ is a $M\times M\times M$ tensor
  • $D$ is a diagonal $2\times 2$ matrix

The entries of the matrices and the tensor are probabilities so the result should somehow be the consequence of Bayes formula. The problem is that I have no idea how to compute that because I don't know how to use tensors. I had an algebra course about tensor products of vector spaces a long time ago but it was very abstract so I don't know how to multiply tensors in practice.

I'm surprised because the first matrix of the product has $2$ rows, the last one has $M$ columns and yet the result is a $M\times M$ matrix.

Could you explain how to do this? For example, what's the dimension of $OT$? I am familiar with the Kronecker product of matrices, is it useful here?


Stupid me... I've spent hours trying to understand this product and... this was a typo. It was $O^TDO$ and the equality was straightforward... I've been confused by the fact that the tensor $T$ did exist and there could be and equality involving it. At least I've learnt a few things about tensors, thank you again for the answers!

  • $\begingroup$ it seems that the following link may help math.stackexchange.com/questions/1133497/…. $\endgroup$ Oct 4, 2016 at 12:27
  • $\begingroup$ If you use the n-mode product as in the link above, then: 1) $OT$ can have dimensions $2\times M\times M$ or $M\times 2\times M$ or $M\times M\times 2$; 2) $R=OTDO$ cannot be a matrix (i.e., to be an $M\times M\times 1$ or $M\times 1\times M$ or $1\times M\times M$) tensor. $\endgroup$ Oct 4, 2016 at 12:30
  • $\begingroup$ So this product is not well defined because there is not a unique way to interpret it, and because the dimensions do not match, right? $\endgroup$
    – Augustin
    Oct 4, 2016 at 12:41
  • $\begingroup$ Yes, it can, for instance, be like this $R=T\times_1 O\times_1 D\times_3 O=T\times_1 DO\times_3 O$, which change dimensions as follows $M\times M\times M\rightarrow 2\times M\times M\rightarrow 2\times M\times 2$ $\endgroup$ Oct 4, 2016 at 12:47

1 Answer 1


The multiplication of a tensor by a matrix (or by a vector) is called $n$-mode product.

Let $\mathcal{T} \in \mathbb{R}^{I_1 \times I_2 \times \cdots \times I_N}$ be an $N$-order tensor and $\mathbf{M} \in \mathbb{R}^{J \times I_n}$ be a matrix. The $n$-mode product is defined as

$$ (\mathcal{T} \times_{n} \mathbf{M})_{i_{1}\cdots i_{n-1}ji_{n+1}\cdots i_N} = \sum \limits_{i_n = 1}^{I_n} \mathcal{T}_{i_{1}i_{2}\cdots i_{n}\cdots i_{N}}\mathbf{M}_{ji_{n}}.$$

Note this is not a standard product like the product of matrices. However, you could perform a matricization of the tensor along its $n$-mode (dimension $n$) and thus effectuate a standard multiplication.

The $n$-mode matricization of $\mathcal{T}$, say $\mathbf{T}_{(n)}$, is an $I_{n} \times I_{1}\cdots I_{n-1}I_{n+1}\cdots I_{N}$ matrix representation of $\mathcal{T}$. In other words, it is just a matrix form to organize all the entries of $\mathcal{T}.$ Hence, the multiplications below are equivalent

$$\mathcal{Y} = \mathcal{T} \times_{n} \mathbf{M} \iff \mathbf{Y}_{(n)} = \mathbf{M} \mathbf{T}_{(n)}, $$ where $\mathbf{Y}_{(n)}$ is the $n$-mode matricization of the tensor $\mathcal{Y} \in \mathbb{R}^{I_1 \times \cdots \times I_{n-1} \times J \times I_{n+1} \times \cdots \times I_N}$.

For more details, see Tensor Decompositions and Applications.


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