Why are the minimum distances of a code and its interleaved code equal?

Question

I know this is probably a very trivial question, but I am completely stuck. Let $C$ be a linear $[n,k,d]$ Code. Then the interleaving of depth $t$ is the Code $C(t)=\{(c_{11}, \dots, c_{t1}, \dots, c_{1n}, \dots, c_{tn}) \mid (c_{i1}, \dots, c_{in}) \in C \}$. Now I am trying to understand why the minimum distance of $C(t)$ is also $d$.

As far as I understand it, $C(t)$ is also linear. Hence minimum weight equals minimum distance. But (as the weights of codewords in $C(t)$ are sums of weights of several codewords in $C$) the minimum weight of codewords in $C(t)$ is obviously larger than of those in $C$. So how can the minimum distance of $C(t)$ be $d$, as well? Any explanation would be very much appreciated.

Answer 1

Consider the distance between the two interleaved codewords formed from codewords $\mathbf c_1, \mathbf c_2, \ldots, \mathbf c_t \in C$ and $\,\mathbf 0, \mathbf c_2, \ldots, \mathbf c_t \in C$ where $\mathbf c_1$ is a codeword of minimum weight $d$. The distance is $d$, no?

Answer 2

You may have misunderstood "minimum distance". It's the smallest Hamming distance of all the pairs of codewords in the code. The Hamming distance is the same for any pair of codewords regardless of how you order the bits/symbols so the minimum distance of the code is unchanged.