Suppose we have a random variable $X$ taking in values in $\{ x_1, .., x_a \}$, and a discrete memoryless channel (DMC), $M$, represented as a matrix, with output alphabet $\{ y_1, .., y_b\}$where its entries $M_{xy} = p_{Y|X}(y|x)$ are the conditional probabilities of the output.
The question now asks to
Show: if two columns of $M$ are linearly dependent, say (WLOG) $\mathbf c_i = k\mathbf c_j$ for a constant $k$, then the channel $M''$ obtained via merging the columns $\mathbf c_i$ and $\mathbf c_j$ by summing them together (so $M''$ has 1 fewer column than $M$) has the same capacity as $M$.
Here capacity $C$ of a DMC is defined as $C = sup\ I(X;Y)$ over the distributions $p_X$ of $X$.
Any help is greatly appreciated!
The idea here is that, to decode, it makes no difference to know that the output is $y_i$, $y_j$, hence it's equivalent to fuse the two ouputs in one. With that idea, consider
$$ H(X|Y) = \sum_y p(Y=y) H(X|Y=y) \tag1$$
Separate the two terms in the sum with the two elements $y_i$, $y_j$:
$$p(Y=y_i) H(X|Y=y_i)+p(Y=y_j) H(X|Y=y_j) \tag2$$
Now, using Bayes inversion
$$ \frac{p(X=x_k|Y=y_j)}{p(X=x_k|Y=y_i)} = c \frac{P(Y=y_i)}{P(Y=y_j)} \tag3 $$ which does not depend on $k$, hence, as expected. This readily implies $p(X=x_k|Y=y_j)=p(X=x_k|Y=y_i)$ and $H(X|Y=y_i)=H(X|Y=y_j)$
Can you go on from here?