In this case, you seem to be interested in the association between two documents' word frequencies, without specifying which is supposedly an "outcome" variable as would be required when using linear regression.
Your best, and simplest bet for measuring the association between two word frequencies is a correlation coefficient. This works whether you have reduced the words to relative frequencies within texts or counted word occurrence as a binary variable. In case of the latter, Pearson's R is equivalent to something called the phi coefficient, with largely the same interpretation (but with a maximum value determined by the relative distribution of your 1s and 0s).
I note an ambiguity in your question, which is: Why are the values in your columns (apparently) proportions if your word frequency is, as you state in the text, 1 and 0 for occurrence or non-occurrence? 1/0 values are word frequency, it's a binary measure of word occurrence. As you will see from my example below, it matters in what sort of answer you get (since the questions are different). It does concern me a bit that your values in the example are not even relative word frequency, however, since they do not (cannot) sum to 1.0 across rows or even columns.
To demonstrate, I will use the quanteda package from R. Here I had to coerce the "document-feature matrix" object into a correlation for the ifelse and cor functions to work.
> require(quanteda)
> myDfm <- dfm(inaugTexts, stem = TRUE)
Creating a dfm from a character vector ...
... lowercasing
... tokenizing
... indexing documents: 57 documents
... indexing features: 9,214 feature types
... stemming features (English), trimmed 3793 feature variants
... created a 57 x 5421 sparse dfm
... complete.
Elapsed time: 0.192 seconds.
>
> wordMatCounts <- as.matrix(myDfm[, c("state", "citizen")])
> wordMatCounts[1:10, ]
features
docs state citizen
1789-Washington 2 4
1793-Washington 0 1
1797-Adams 12 3
1801-Jefferson 3 5
1805-Jefferson 12 10
1809-Madison 5 0
1813-Madison 5 3
1817-Monroe 29 9
1821-Monroe 27 14
1825-Adams 8 3
>
> wordMatRelFreq <- as.matrix(weight(myDfm, "relFreq")[, c("state", "citizen")])
> wordMatBinary <- ifelse(as.matrix(wordMatCounts) > 0, 1, 0)
>
> cor(wordMatCounts)
state citizen
state 1.0000000 0.6338093
citizen 0.6338093 1.0000000
> cor(wordMatRelFreq)
state citizen
state 1.00000000 0.03351566
citizen 0.03351566 1.00000000
> cor(wordMatBinary)
state citizen
state 1.0000000 0.2668415
citizen 0.2668415 1.0000000
