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Consider the matrix product $A=B\cdot C$, where $A$ is a $n\times n$ matrix, $B$ is a $n\times m$ matrix ($n<m$), and $C$ is a $m\times n$ matrix.

Suppose that $\rho(B)=n$ (i.e., $B$ has full row rank).

My question is: under what condition (over $C$) can we say the rank of $A$ is also $n$ ($A$ is invertible)?

Or, at least, is it possible to show that for a fixed $B$, there exists a $C$ such that the product preserves the rank?

  1. This ($\rho(A)=n$) is possible since it does not violate the rank inequity $\min\{\rho(B),\rho(C)\}\geq \rho(A)\geq \rho(B)+\rho(C)-m$.

  2. A necessary (but not sufficient) condition is $\rho(C)=n$ (by the above inequity). Therefore, I am expecting a stronger requirement of $C$ than just having full col. rank.

Mingzhou Liu
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  • Do you have example, that 2 point is not sufficient? – zkutch Nov 30 '23 at 11:19
  • Very simple, consider $B=\begin{bmatrix} 1 & 0 & 0 & 0\ 0 & 1 & 0 & 0 \end{bmatrix}$ and $C=\begin{bmatrix} 0 & 0\ 0 & 0\ 1 & 0\ 0 & 1 \end{bmatrix}$, we have $A=[0]$. – Mingzhou Liu Nov 30 '23 at 11:30

1 Answers1

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Edit: I sat on the question a bit, I think I came up with a full answer. I will leave my old answer below.

Based on my old answer, it should be intuitive that, if the image of $C$ does not contain any vector in the kernel of $B$, then $A$ is full rank. I make this idea rigorous in the following proposition.

Proposition. If $B$ and $C$ are full row and full column rank, respectively, then $A$ is full rank if and only if $\ker(B) \cap \mathrm{im}(C) = \emptyset$, where $\ker(B)$ is the kernel of $B$ and $\mathrm{im}(C)$ is the image of $C$.

Proof. The "if" direction is trivial: for any $x\in\mathbb{R}^n\backslash\{0\}$, we have that $Cx = y\in\mathrm{im}(C)$, meaning that $By = BCx = Ax \neq 0$, for all $x\neq0$. We prove the "only if" direction by contrapositive. Suppose that there exists $y\neq 0$ such that $y\in\ker(B) \cap \mathrm{im}(C)$. Since it is in the image of $C$, there exists $x\in\mathbb{R}^n\backslash\{0\}$ such that $y=Cx$. This implies $Ax = 0$, i.e., $A$ is rank deficient, concluding the proof.

I think this answers your first question, while the answer to the second one is discussed below.

Old answer. I think your question is complicated to answer in full generality, I will just give you a few pointers. This is not really an answer, it is more a comment which is too long for the comment section.

Regarding the point 2, you are correct, $\rho(C)=n$ is not a sufficient condition. Since the kernel of $B$ is at least of dimension $1$, it is possible to construct a full column rank $C$ so that one of its columns is in $\ker(B)$, meaning that one column of $A$ is only zeros (i.e., $A$ is rank deficient). In fact, as a side note, if $m\geq 2n$ you can easily construct a full column rank $C$ so that $A$ is the zero matrix!

However, regarding your second question, yes, it is always possible to find a matrix $C$ so that, for a given $B$, the product is full rank: it suffices to take $C = B^\top$. Since $\ker(B^\top) = \{0\}$, you have that the product $A=BB^\top$ is not only invertible but positive definite, since, for any vector $x\in\mathbb{R}^n\backslash\{0\}$: $$ x^\top Ax = x^\top BB^\top x = \lVert B^\top x\rVert^2 >0. $$ I hope my comments were helpful.

Carlos Santi Toledo
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  • Thanks a lot for the second pointer! Since I am interested in constructing $C$ to be a transition probability matrix (sum of each col. should be 1), is it possible to let $C=\bar{B^T}$, where the bar denotes averaging each col. w.r.t. col. sum and still have $A=BC$ is invertible? – Mingzhou Liu Nov 30 '23 at 11:54
  • If I understand correctly, you want to take $B^\top$ and normalize each column, is it? If it is the case, then yes, you can do that, since changing the columns' norms does not change the image of the matrix (see my updated answer!). – Carlos Santi Toledo Nov 30 '23 at 12:07