# When is $A^TBA$ invertible, where $B$ is an invertible, symmetric matrix? [closed]

Let $$B$$ be an $$n×n$$ invertible matrix such that $$B^T=B$$, and let $$A$$ be a $$n×2$$ matrix with linearly independent columns. When is the product $$A^TBA$$ invertible?

• Isn't the rank of A smaller than that of B, and thus the product is not invertible?
– Jack
Dec 27, 2021 at 23:24
• A sufficient condition is that $B$ is positive definite or negative definite. But this is not a necessary condition. Dec 28, 2021 at 0:15
• @Jack The product has size $2 \times 2$, so there's no issue there Dec 28, 2021 at 1:19
• Please do not edit savagely your question. Apr 3, 2022 at 18:33

We'll prove $$A^TBA$$ is invertible $$\iff$$ $$B\left(\text{Col}(A)\right)\cap \text{Col}(A)^{\perp}=\{0\}$$.
Here is the "only if" part. Assume $$A^TBA$$ is invertible but there is some non$$-$$zero $$y$$ in $$B\left(\text{Col}(A)\right)\cap \text{Col}(A)^{\perp}$$. Then $$y=BAx$$ for some $$x\in \mathbb{R}^2$$. Because $$\text{Col}(A)^{\perp}=\text{Nul}(A^T)$$ we also get $$A^Ty=0$$. Putting both together implies $$A^TBAx=0$$ i.e. $$x\in \text{Nul}(A^TBA)$$. Now $$A^TBA$$ is assumed to be invertible, so $$x$$ must be the zero vector, and so is $$y$$, a contradiction.
Here is the "if" part. Assume $$B\left(\text{Col}(A)\right)\cap \text{Col}(A)^{\perp}$$ has a trivial intersection and $$A^TBAx=0$$ for some $$x\in \mathbb{R}^2$$. Then we have $$BAx\in \text{Nul}(A^T)=\text{Col}(A)^{\perp}$$. But $$BAx$$ also belongs to $$B\left(\text{Col}(A)\right)$$ which means $$BAx=0$$ from our hypothesis. Hence $$Ax=B^{-1}0=0$$ and since $$A$$ has independent columns, $$x=0$$ and $$A^TBA$$ is invertible.
A slightly different way to look at this same idea: $$A^TBA$$ is invertible exactly if it has rank 2. Trivally we have that $$BA$$ has rank 2. Thus $$BA$$ can be understood as a map that maps the two canonical unit vectors to $$Ba_1$$ and $$Ba_2$$ where $$a_i$$ are the columns of $$A$$.
So $$A^TBA$$ is invertible exactly if $$A^TBa_1$$ and $$A^TBa_2$$ are independent. This means $$(a_1Ba_1, a_2Ba_1)$$, $$(a_1Ba_2, a_2Ba_2)$$ need to be independent. Note that as $$B$$ is symmetric this means $$(a_1Ba_1, a_1Ba_2)$$, $$(a_1Ba_2, a_2Ba_2)$$ need to be independent. This is equivalent by using the determinant to $$(a_1Ba_1)(a_2Ba_2) - (a_1Ba_2)^2 \neq 0$$, or equivalently to $$(a_1Ba_1)(a_2Ba_2) \neq (a_1 B a_2)^2$$
EDIT: Note that if $$B$$ is positive definite the above is related to Cauchy-Schwarz. In this case $$xBy$$ defines a scalar product, so $$(a_1Ba_2)^2 \leq (a_1Ba_1)(a_2Ba_2)$$ and equality only if $$a_1,a_2$$ are linearly dependent. So in this case $$a_1,a_2$$ independent already implies this property.