Why is $BB^T$ always invertible?

Question

In Karmarkar’s method, we use $$[I - B^T(BB^T)^{-1}B]v$$ Why does $BB^T$ always have an inverse?

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Karmarkar’s method is applied to an LP in the following form:

$\min z = cx$

subject to

$AX=0$

$x_1 +x_2 +......+ x_n =1$

$X\ge0$

$x =[x_1 ,x_2,.....,x_n]^T$, $A$ is an $m \times n$ matrix, $c = [c_1, c_2, ..... ,c_n]$ ,and 0 is an n-dimensional column vector of zeros. The LP must also satisfy $[\frac{1}{n},\frac{1}{n},.....,\frac{1}{n}]^T$ is feasible , Optimal $z-$value $=0$

B is the $(m * 1) * n$ matrix whose first m rows are A and whose last row is a vector of $1’$s.

$B = \begin{bmatrix}A\\1 \end{bmatrix}$

Answer 1

The statement is not true.

For example if $A= \begin{bmatrix} 2& 3\\1 & 1 \end{bmatrix}$,then $det(BB^T)=0$ so $BB^T$ is not invertible.