Let $A$ be a symmetric $n \times n$ matrix, and $V = \mathbb{R}^n$. Define $\langle v,w \rangle = v^t A w$ with $v,w \in V$. Show that $A$ is positive definite iff $\det(A_k) > 0$ for $1 \leq k \leq n$ where $A_k$ is the $k \times k$-matrix which is the left upper $k \times k$ corner of A.
If $k=1$ it was quite easy. I'm having troubles with the rest. I tried using the fact that the determinant of a matrix is the product of its eigenvalues but I didn't get anywhere.
$(\Leftarrow)$ We proceed by induction. You have already proven the base case. Let us assume that it holds for all $(n - 1) \times (n - 1)$ matrices $B$ with all $\Delta_k = \det(B_k)$, where $1 \leq k \leq ( n -1)$. (Here I'm using the same notation as you: $B_k$ is the upper-left $k \times k$ matrix of $B$). Let us further assume that we are now dealing with an $n \times n$ matrix $A$ whose upper left $(n - 1) \times (n - 1)$ matrix is $B$.
The goal of the inductive step will be to show that $A$ as described has no non-positive eigenvalues, which is a necessary and sufficient condition for being positive definite. Our first aim in the inductive step will be to show that there is at most one non-positive eigenvalue (counting multiplicites) of $A$. That is, there is at most one non-positive eigenvalue and it has multiplicity at most $1$. Indeed, suppose the contrary: if there existed either two distinct non-positive eigenvalues or a non-positive eigenvalue with multiplicity at least $2$, then by the Spectral Theorem we could find two eigenvectors $\vec{u}$ and $\vec{v}$ such that $A \vec{u} = \lambda_1 \vec{u}$, $A \vec{v} = \lambda_2 \vec{v}$, $\|\vec{u}\| = \|\vec{v}\| = 1$ and $\vec{u} \cdot \vec{v} = 0$. (Note here that $\lambda_1, \lambda_2 \leq 0$ and it could be that $\lambda_1 = \lambda_2$.) Let us choose $\vec{w}$ to be a non-trivial linear combination of $\vec{u}$ and $\vec{v}$ whose last entry is zero. (For example, we can choose $\vec{w} = v_n \vec{u} - u_n \vec{v}$, where $v_n$ and $u_n$ are the $n$th entries of $\vec{u}$ and $\vec{v}$, respectively.
On one hand, $$\vec{w}^T A \vec{w} = v_n^2 \vec{u}^T A \vec{u} + u_n^2 \vec{v}^T A \vec{v} = \lambda_1 v_n^2 + \lambda_2 u_n^2 \leq 0$$ However, since $w_n = 0$, it is the case that $$\vec{w}^T A \vec{w} = \vec{x}^T B \vec{x} > 0$$ where $\vec{x}$ is $\vec{w}$ with the last entry truncated. Note that the second inequality follows from the inductive hypothesis, as $B$ is assumed to be positive-definite. We have reached a contradiction, so indeed there is at most one non-positive eigenvalue, including multiplicities.
However, it is not possible to have a single eigenvalue with multiplicity one be non-positive. Indeed, the determinant of $A$, which is assumed to be positive, is the product of the eigenvalues (counting multiplicities). If this were the case, then $\det A$ would have to be non-positive, as the product of $n - 1$ positive eigenvalues and the singular non-positive eigenvalue (recall we've shown that at least $n - 1$ of $A$'s eigenvalues are positive). The claim is proven. $\square$
$(\Rightarrow)$ To prove necessity, we will prove the contrapositive. Let $k^*$ be the minimal value of $k$ such that $\Delta_k \leq 0$. If we retrace the steps of the proof of the previous claim, then we can note that $A_k$ for all $k < k^*$ is positive definite, but since $\Delta_{k^*} \leq 0$, it follows that $A_{k^*}$ has at least one non-positive eigenvalue. But if $A_{k^*}$ is not positive definite, then so too is $A$. To see why this is, let $\vec{v} \in \mathbb{R}_{k^*}$ be the vector such that $\vec{v}^T A_{k^*} \vec{v} \leq 0$. Then by filling $\vec{v}$ with zeroes until it has $n$ entries to obtain the vector $\vec{w}$, we have obtained a vector $\vec{w}$ such that $\vec{w}^T A \vec{w} \leq 0$. $\square$
