If $\mathbf A$ is any square non-singular matrix of dimension $n \times n$. And $\mathbf B$ is a $n \times m$ matrix with $\mathrm{rank(\mathbf B)} = m$. Is the full rank condition of matrix $\mathbf B$ both sufficient as well as necessary to state that the matrix product $\mathbf {B^TAB}$ is non-singular?

i.e., can we write:

$ \mathbf {A}$ is non-singular $\iff$ $\mathrm{rank(\mathbf B)} = m$ and $\mathbf {B^TAB}$ is non-singular

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    $\begingroup$ The answer seems to be negative. There are examples that $A$ is non-singular and $\mathrm{rank}(B)=m$, but $B^TAB=0$. $\endgroup$ – Quang Hoang Aug 18 '14 at 13:16

Nope, the statement is not true in this form. A simple example is: $$ B=\begin{bmatrix}1\\1\end{bmatrix},\quad A=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Then $\mathrm{rank}(B)=1$ and $B^TAB=1$ is non-singular, but $A$ is singular. If $$ B=\begin{bmatrix}1\\1\end{bmatrix}, \quad A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}, $$ then $\mathrm{rank}(B)=1$, $A$ is non-singular, but $B^TAB=0$ is singular.

The condition $B^TAB$ and $\mathrm{rank}B=m$ does not need to be sufficient for the non-singularity of $A$ since the conditions impose a sort of "non-singularity on a restricted subspace", which obviously does not need to be sufficient for the "non-singularity of the whole". On the other hand, a non-singular $A$ does not imply that $x^TAx\neq 0$ (e.g., indefinite symmetric matrices).

NOTE: $\mathrm{rank}(B)=m$ should actually be an assumption for the equivalence (if there was any) as the non-singularity of $A$ and the full rank of $B$ are completely independent things (non-singularity of $A$ does not imply anything about the rank of $B$).

  • $\begingroup$ So, the required condition could be: $\mathbf B$ to be a matrix of dimension $n\times n$ with $\mathrm{rank}(\mathbf B)=n$. $\endgroup$ – neelarnab Aug 19 '14 at 12:59
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    $\begingroup$ @neelarnab In that case, it is of course true but with $\mathrm{rank}(B)=n$ being an assumption for the equivalence: "Let $B\in\mathbb{R}^{n\times n}$, $\mathrm{rank}(B)=n$. Then $A$ is non-singular iff $B^TAB$ is non-singular". $\endgroup$ – Algebraic Pavel Aug 19 '14 at 13:17

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