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When you multiply a matrix $M$ by its transpose, what exactly does this product represent? What does each entry represent?

I see that a lot of these examples, when a document-term matrix (DTM) is created and then this DTM is multiplied by its transpose and they call it the term-term matrix.

Rodrigo de Azevedo
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sanjeev
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2 Answers2

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If the rows of $M$ are $m_1, m_2,\ldots, m_n$, then the $i,j^\textrm{th}$ entry of $MM^T$ is $m_i\cdot m_j$, the dot product.

vadim123
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The matrix product $X=A^t*A$ is essential in several contexts. If you are only looking for an interpretation of its elements then the most obvious is that $X$ contains the cross-correlations of the columns of $A$. $X$ is always square, symmetric, and positive semi-definite (all eigenvalues are >= 0). The diagonal values contain the sum of the squares of each column, and a sum can never be zero unless a column of $A$ contains all zeros. The off-diagonal values of $X$ are the cross-correlations between columns. For example, with

$ A=\begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6\\ \end{bmatrix} $ , $ X=\begin{bmatrix} 14 & 32\\ 32 & 77\\ \end{bmatrix} $ .

The eigenvalues of $X$ are both greater than zero because the columns of $A$ are linearly independent. On the other hand, with

$ A=\begin{bmatrix} 1 & 2\\ 2 & 4\\ 3 & 6\\ \end{bmatrix} $ , $ X=\begin{bmatrix} 14 & 28\\ 28 & 56\\ \end{bmatrix} $

which is singular because the columns of $A$ are linearly dependent (eigenvalues 0 and 70). $X$ always has the same rank as $A$ regardless of the number of rows and columns in A. The off-diagonal elements in $X$ can be zero if two columns in $A$ are orthogonal. As an example, with

$ A=\begin{bmatrix} 1 & 3\\ 2 & 0\\ 3 & -1\\ \end{bmatrix} $ , $ X=\begin{bmatrix} 14 & 0\\ 0 & 10\\ \end{bmatrix} $ .

If you have more than a passing interest in the subject, have a look at https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse and https://en.wikipedia.org/wiki/Singular_value_decomposition. It is well worth the effort.

olekirkeby
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