Invertibility of the Gram matrix of a convex combination Let's assume two real valued matrices $A,B\in R^{w\times d}$ for which $d>w$ and they both have full (column) rank.
I am interested in the invertibility of the Gram matrix $$H(t):=(A+t(B-A))(A+t(B-A))^T$$ for $t\in(0,1)$ (note that $A+t(B-A)$ is the line segment between $A$ and $B$).
If I am correct, there are infinitely many $t$ that $H(t)$ is invertible.
My question is, for which $t$ is $H(t)$ not invertible? Are there finitely many such $t$? I guess based on the polynomial properties of the determinant, it should be finite, right? If yes, can any one help me to prove that?

Introducing the Gram matrix complicates the problem (and makes computational study of it difficult.)  Instead, notice that $H(t)$ is invertible exactly when $A + t(B-A) = A + tC$ is of full rank.
We can study this in detail in terms of the Singular Value Decomposition (SVD) of $A,$
$$A = UDV^\prime,$$
where $U^\prime U = \mathbb{I}_d = VV^\prime$ and $D$ is the diagonal matrix of the $d$ singular values of $A,$ all of which are nonzero because $A$ is of full rank.  Accordingly, both $U$ and $V$ are of full rank, so instead let's examine the rank of the $d\times d$ matrix
$$U^\prime H(t) V = D + tU^\prime C V = tD\left(t^{-1}\mathbb{I}_d + D^{-1}U^\prime C V\right),$$
which will be the same as that of $H(t).$  (The rank of $H(0)$ is, of course, the rank of $A,$ so from now on assume $t\ne 0.$)
Taking determinants and writing $p_X(\lambda) = \det(X - \lambda \mathbb I)$ for the characteristic polynomial of any matrix $X,$
$$\begin{aligned}
\det \left(U^\prime H(t)V\right) &= \det(tD)\det\left(t^{-1}\mathbb{I}_d + D^{-1}U^\prime C V\right)\\
&=\det(D)\,t^d\ p_{D^{-1}U^\prime (B-A) V}(-t^{-1}).
\end{aligned}$$
Since $\det(D)\ne 0$ and $t\ne 0,$ the $t$ for which $H(t)$ is singular are the roots of $p_X(-t^{-1})$ where $X = D^{-1}U^\prime (B-A) V.$
Since the roots of $p_X$ are the eigenvalues of $X,$ we have found that

The only $t$ for which $H(t)$ fails to be invertible are the negative reciprocals of the (nonzero) eigenvalues of $X.$

(This includes complex eigenvalues, BTW.)
Since any $d\times d$ matrix has at most $d$ distinct eigenvalues, there are only finitely many such $t$ -- and we now have a means to compute them all.  The ones relevant to the question are the real eigenvalues in the interval $(-\infty, -1),$ which may be fewer in number.
Example
Here is a simple but non-trivial example with $w\times d = 3\times 2$ matrices.  Let
$$A = \pmatrix{1&0\\0&1\\0&0},$$
which obviously is of full rank, and
$$B = \pmatrix{-1&-1\\-1&-1\\0&0}.$$
(Although the question stipulates that $B$ should be of full rank, nothing in the foregoing analysis requires that and this example helps show why that is unnecessary.)
Here is R code to assess the invertibility of $H(t)$ by computing its determinant:
detH <- Vectorize(function(x, A, B) det(crossprod(A + x*(B - A))), "x")

I vectorized the calculation to enable easy plotting of $\det (H(t)),$ as in this snippet and its output:
A <- rbind(diag(1,2), 0)
B <- rbind(matrix(-1,2,2), 0)
curve(detH(t, A, B), 0, 1, xname="t", n=501, lwd=2)
abline(v = c(1/3, 1), col="Red", lty=3)


Evidently $H(t)$ fails to be of invertible precisely when $t\in\{1/3,1\},$ because those are the two points where this graph touches $0.$
What does the analysis say?  We must begin with the SVD of $A.$ $A$ already is column-orthogonal, meaning $A=U,$ whence $D=V=\mathbb{I}_2$ are both identity matrices.  Thus
$$X = D^{-1}U^\prime (B-A) V = U^\prime (B-A) = \pmatrix{1&0&0\\0&1&0}\pmatrix{-2&-1\\-1&-2\\0&0}=\pmatrix{-2&-1\\-1&-2}.$$
Notice that although $A$ and $B$ are rectangular matrices, $X$ is a square $d\times d$ matrix.  Its characteristic polynomial is
$$p_X(\lambda) = \det(X - \lambda\mathbb{I}_2) = \det\pmatrix{-2-\lambda&-1\\-1&-2-\lambda}=(2+\lambda)^2-1.$$
Its zeros all satisfy $(2+\lambda)^2=1,$ whence the eigenvalues are $\lambda\in\{-3,-1\}.$  Their negative reciprocals are $t\in\{1/3,1\},$ agreeing with the direct R calculation.
