# Rank of concatenated matrix

Given $N$ matrices $X_k=A_k^T\otimes (B_k\cdot C_0)$ and the vertically concatenated matrix $X = [X_1^T\; X_2^T\; \dots \;X_N^T]^T$, what is the condition for $X$ to have full rank? The matrices $A_k,\;B_k$ and $C_0$ have all full rank and appropriate dimensions. The operator $\otimes$ denotes the krnoecker product.

The rank of $X_k$ can be expressed as $\operatorname{rank}(A_k^T\otimes (B_k\cdot C_0))= \operatorname{rank}(A_k)\cdot \operatorname{rank}(B_k\cdot C_0)$. But what about the rank of the concatenated matrix $X$?

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The row space of $X = \pmatrix{X_1\cr X_2\cr \ldots\cr X_N\cr}$ is the span of the rows of all the $X_i$. If $X$ is $m \times n$, the condition for $X$ to have full rank is that there are $\min(m,n)$ linearly independent rows.
Thanks for this answer. I'm wondering if there are more precise conditions for $A_k$, $B_k$ and $C_0$ as they form the matrix $X$ in a very special way. – mburg Jul 4 '12 at 7:28
If $A_k^T$ is $m_k \times n$ and $B_k . C_0$ is $p_k \times r$, $A_k^T \otimes (B_k \cdot C_0)$ is $(m_k p_k) \times (n r)$, and its rows are the tensor products of the rows of $A_k^T$ and $B_k \cdot C_0$. – Robert Israel Jul 4 '12 at 17:16