[proofplan] We use the definition of linear dependence to obtain a nontrivial relation $\sum c_i f_i = 0$ on all of $I$. Differentiating this identity $0, 1, \ldots, k-1$ times produces a homogeneous linear system whose coefficient matrix at each point $t$ is exactly the Wronskian matrix. Since the system has a nontrivial solution $(c_1, \ldots, c_k)$, the determinant of the Wronskian matrix — the Wronskian $W(f_1, \ldots, f_k)(t)$ — must vanish at every $t \in I$. [/proofplan]

[step:Define the Wronskian matrix and recall the setup] The Wronskian matrix of functions $f_1, \ldots, f_k \in C^{k-1}(I; \mathbb{R})$ at a point $t \in I$ is the $k \times k$ matrix \begin{align*} W(t) := \begin{pmatrix} f_1(t) & f_2(t) & \cdots & f_k(t) \\ f_1'(t) & f_2'(t) & \cdots & f_k'(t) \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(k-1)}(t) & f_2^{(k-1)}(t) & \cdots & f_k^{(k-1)}(t) \end{pmatrix} \in \mathbb{R}^{k \times k}, \end{align*} where the entry in row $r$ and column $j$ is $f_j^{(r-1)}(t)$ for $r = 1, \ldots, k$ and $j = 1, \ldots, k$. The derivatives $f_j^{(r-1)}$ exist and are continuous because $f_j \in C^{k-1}(I; \mathbb{R})$. The Wronskian is the determinant: \begin{align*} W(f_1, \ldots, f_k)(t) := \det W(t). \end{align*} [/step]

[step:Differentiate the dependence relation to build a homogeneous system]Since $(f_1, \ldots, f_k)$ is linearly dependent in $C^{k-1}(I; \mathbb{R})$, there exist scalars $c_1, \ldots, c_k \in \mathbb{R}$, not all zero, such that \begin{align*} c_1 f_1(t) + c_2 f_2(t) + \cdots + c_k f_k(t) = 0 \quad \text{for all } t \in I. \end{align*} This is an identity of functions on $I$. Since each $f_j \in C^{k-1}(I; \mathbb{R})$, the left-hand side is $(k-1)$-times differentiable. Differentiating the identity successively $1, 2, \ldots, k-1$ times with respect to $t$, the constants $c_j$ pass through each derivative: \begin{align*} c_1 f_1'(t) + c_2 f_2'(t) + \cdots + c_k f_k'(t) &= 0, \\ c_1 f_1''(t) + c_2 f_2''(t) + \cdots + c_k f_k''(t) &= 0, \\ &\;\;\vdots \\ c_1 f_1^{(k-1)}(t) + c_2 f_2^{(k-1)}(t) + \cdots + c_k f_k^{(k-1)}(t) &= 0. \end{align*} Including the original relation (the zeroth derivative), we have $k$ equations, all valid for every $t \in I$.[/step]

[guided]Since $(f_1, \ldots, f_k)$ is linearly dependent in $C^{k-1}(I; \mathbb{R})$, there exist scalars $c_1, \ldots, c_k \in \mathbb{R}$, not all zero, such that \begin{align*} c_1 f_1(t) + c_2 f_2(t) + \cdots + c_k f_k(t) = 0 \quad \text{for all } t \in I. \end{align*} Define the function $g: I \to \mathbb{R}$ by $g := c_1 f_1 + \cdots + c_k f_k$. The equation above says $g$ is the zero function on $I$. Why does this let us differentiate? Because each $f_j \in C^{k-1}(I; \mathbb{R})$, so $g \in C^{k-1}(I; \mathbb{R})$ as well, and all derivatives of $g$ through order $k-1$ exist, are continuous, and are identically zero on $I$. Now we use the linearity of differentiation: for each derivative order $r = 0, 1, \ldots, k-1$, \begin{align*} g^{(r)}(t) = c_1 f_1^{(r)}(t) + c_2 f_2^{(r)}(t) + \cdots + c_k f_k^{(r)}(t) = 0 \quad \text{for all } t \in I. \end{align*} The constants $c_j$ pass through each derivative because they do not depend on $t$. Writing out the full system of $k$ equations explicitly: \begin{align*} c_1 f_1(t) + c_2 f_2(t) + \cdots + c_k f_k(t) &= 0, \\ c_1 f_1'(t) + c_2 f_2'(t) + \cdots + c_k f_k'(t) &= 0, \\ c_1 f_1''(t) + c_2 f_2''(t) + \cdots + c_k f_k''(t) &= 0, \\ &\;\;\vdots \\ c_1 f_1^{(k-1)}(t) + c_2 f_2^{(k-1)}(t) + \cdots + c_k f_k^{(k-1)}(t) &= 0. \end{align*} This gives $k$ equations (one for each derivative order $r = 0, \ldots, k-1$), all holding simultaneously at every $t \in I$, with the same coefficients $(c_1, \ldots, c_k)$. The key observation is that the scalars $c_j$ are constants — they do not depend on $t$ or on the derivative order. This is precisely what will let us assemble these equations into a single matrix equation in the next step.[/guided]

[step:Recognize the system as $W(t) \, c = 0$ and conclude the Wronskian vanishes]Fix any $t \in I$. The $k$ equations from the previous step can be written in matrix form as \begin{align*} W(t) \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \end{align*} where $W(t)$ is the Wronskian matrix defined in the first step. The vector $c = (c_1, \ldots, c_k)^\top$ is nonzero (not all $c_j$ are zero). Therefore the homogeneous system $W(t) \, c = 0$ has a nontrivial solution. A square matrix with a nontrivial null vector has zero determinant: if $\det W(t) \neq 0$, then $W(t)$ would be invertible and the unique solution to $W(t) \, c = 0$ would be $c = 0$, contradicting $c \neq 0$. Therefore \begin{align*} W(f_1, \ldots, f_k)(t) = \det W(t) = 0. \end{align*} Since $t \in I$ was arbitrary, $W(f_1, \ldots, f_k)(t) = 0$ for all $t \in I$.[/step]

[guided]Fix any $t \in I$. The $k$ equations obtained from differentiating the dependence relation (one for each derivative order $r = 0, 1, \ldots, k-1$) can be assembled into matrix form. The equation for derivative order $r$ reads $\sum_{j=1}^k c_j f_j^{(r)}(t) = 0$, which is exactly the $r$-th row of the matrix equation \begin{align*} W(t) \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \end{align*} where $W(t)$ is the Wronskian matrix defined in the first step. The vector $c = (c_1, \ldots, c_k)^\top$ is nonzero (not all $c_j$ are zero, by the definition of linear dependence). Therefore the homogeneous system $W(t) \, c = 0$ has a nontrivial solution, meaning $c$ lies in the null space of $W(t)$. Why does a nontrivial null vector force the determinant to be zero? Because an invertible matrix has trivial null space: if $\det W(t) \neq 0$, then $W(t) \in GL_k(\mathbb{R})$, and the unique solution to $W(t) \, c = 0$ would be $c = W(t)^{-1} \cdot 0 = 0$. This contradicts $c \neq 0$. Therefore \begin{align*} W(f_1, \ldots, f_k)(t) = \det W(t) = 0. \end{align*} Since $t \in I$ was arbitrary, $W(f_1, \ldots, f_k)(t) = 0$ for all $t \in I$. Note that this argument uses the same nonzero vector $c$ at every point $t$. The coefficients $c_1, \ldots, c_k$ come from the dependence relation and are independent of $t$, so $c$ is a simultaneous nontrivial null vector for the entire family $\{W(t)\}_{t \in I}$. This is stronger than merely saying the determinant vanishes — it says the null space contains a fixed direction at every point. It is also important to note that the converse does not hold in general: $W(f_1, \ldots, f_k)(t) = 0$ for all $t$ does not imply linear dependence unless additional structure is present (for instance, when the $f_j$ are solutions of a single linear ODE of order $k$, the converse does hold by the existence and uniqueness theorem).[/guided]