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If $n \leq p$, then obviously this is true. For variables $x_1,\dots,x_n$, form the vector $\vec{v} = (x_1,\dots,x_n)^\dagger$, and consider the polynomial $F(x_1,\dots,x_n) = \text{det}(T_1\vec{v},\dots,T_n\vec{v})$. This is a homogeneous degree $n$ polynomial in the variables $x_1,\dots,x_n$ with coefficients in $\mathbb{F}_p$.

Now prove the following lemma: for every integer $m$, every homogeneous polynomial $F(x_0,\dots,x_m)$ of degree $\leq p$ that vanishes identically on $\mathbb{F}_p^m$ is the zero polynomial. This is proved by induction on $m$. For $m=0$, this is trivial. For $m=1$, this is a slight variant of the inhomogeneous analogue. Every inhomogeneous polynomial $F(1,x)$ that vanishes identically on $\mathbb{F}_p$ is a multiple of $x^p-x$. Since the degree is $\leq p$, this means that $F(x_0,x_1) = a(x_1^p-x_1x_0^{p-1})$ for some $a\in \mathbb{F}_p$. Since $F(0,1)$ is zero, also $a$ equals $0$, so that $F$ is the zero polynomial. That establishes the base case of the induction.

By way of induction, assume that $m\geq 2$, and assume the result for $m-1$. Write $F(x_0,\dots,x_m)$ as $$F_{0}(x_0,\dots,x_{m-1})x_m^n + \dots + F_{n-d}(x_0,\dots,x_{m-1})x_m^d + \dots + F_n(x_0,\dots,x_{m-1})x_m^0,$$
with each $F_{n-d}$ a homogeneous polynomial in $x_0,\dots,x_{m-1}$ of degree $n-d \leq p$. Plug in $(0,\dots,0,1)$ to conclude that the constant $F_0$ is zero. For fixed values $(a_0,\dots,a_{m-1})\in \mathbb{F}_p^{m}$ the resulting polynomial $F(a_0,\dots,a_{m-1},x) \in \mathbb{F}_p[x]$ vanishes identically on $\mathbb{F}_p$ by hypothesis. By the same argument as for $m=1$, $F(a_0,\dots,a_{m-1},x) = a(x^p-x)$ for some $a\in \mathbb{F}_p$. Since $F_0$ is zero, in fact $a$ equals $0$. Thus every $F_{m-d}(a_0,\dots,a_{m-1})$ equals $0$ for every $(a_0,\dots,a_{m-1})$ in $\mathbb{F}_p^m$. By the induction hypothesis, this forces every polynomial $F_{m-d}(x_0,\dots,x_{m-1})$ to be the zero polynomial. Therefore also $F(x_0,\dots,x_m)$ is the zero polynomial.

Finally, for $n=p+1$, there is a generalization of Abhinav's counterexample. Namely, the $p+1$ matrices $T_0,\dots,T_p$ defined by, $$ T_0(\sum_{i=0}^px_i \mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i, \ \ T_0 = \text{Id}, $$ and for $i=1,\dots,p$, $$ T_i(\sum_{i=0}^p x_i\mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i + (x_p-ix_0)\mathbf{e}_i, $$ (these are all invertible except if $p=2$, where a shuffle as in Abhinav's answer works). The determinant is $F(x_0,x_1,\dots,x_p) = x_0x_p(x_p-1x_0)\cdots (x_p-(p-1)x_0)$, which vanishes identically on $\mathbb{F}_p^{p+1}$.

If $n \leq p$, then obviously this is true. For variables $x_1,\dots,x_n$, form the vector $\vec{v} = (x_1,\dots,x_n)^\dagger$, and consider the polynomial $F(x_1,\dots,x_n) = \text{det}(T_1\vec{v},\dots,T_n\vec{v})$. This is a homogeneous degree $n$ polynomial in the variables $x_1,\dots,x_n$ with coefficients in $\mathbb{F}_p$.

Now prove the following lemma: for every integer $m\geq 0$, every homogeneous polynomial $F(x_0,\dots,x_m)$ of degree $\leq p$ that vanishes identically on $\mathbb{F}_p^{m+1}$ is the zero polynomial. This is proved by induction on $m$. For $m=0$, this is trivial. For $m=1$, this is a slight variant of the inhomogeneous analogue. Every inhomogeneous polynomial $F(1,x)$ that vanishes identically on $\mathbb{F}_p$ is a multiple of $x^p-x$. Since the degree is $\leq p$, this means that $F(x_0,x_1) = a(x_1^p-x_1x_0^{p-1})$ for some $a\in \mathbb{F}_p$. Since $F(0,1)$ is zero, also $a$ equals $0$, so that $F$ is the zero polynomial. That establishes the base case of the induction.

By way of induction, assume that $m\geq 2$, and assume the result for $m-1$. Write $F(x_0,\dots,x_m)$ as $$F_{0}(x_0,\dots,x_{m-1})x_m^n + \dots + F_{n-d}(x_0,\dots,x_{m-1})x_m^d + \dots + F_n(x_0,\dots,x_{m-1})x_m^0,$$
with each $F_{n-d}$ a homogeneous polynomial in $x_0,\dots,x_{m-1}$ of degree $n-d \leq p$. Plug in $(0,\dots,0,1)$ to conclude that the constant $F_0$ is zero. For fixed values $(a_0,\dots,a_{m-1})\in \mathbb{F}_p^{m}$ the resulting polynomial $F(a_0,\dots,a_{m-1},x) \in \mathbb{F}_p[x]$ vanishes identically on $\mathbb{F}_p$ by hypothesis. By the same argument as for $m=1$, $F(a_0,\dots,a_{m-1},x) = a(x^p-x)$ for some $a\in \mathbb{F}_p$. Since $F_0$ is zero, in fact $a$ equals $0$. Thus every $F_{m-d}(a_0,\dots,a_{m-1})$ equals $0$ for every $(a_0,\dots,a_{m-1})$ in $\mathbb{F}_p^m$. By the induction hypothesis, this forces every polynomial $F_{m-d}(x_0,\dots,x_{m-1})$ to be the zero polynomial. Therefore also $F(x_0,\dots,x_m)$ is the zero polynomial.

Finally, for $n=p+1$, there is a generalization of Abhinav's counterexample. Namely, the $p+1$ matrices $T_0,\dots,T_p$ defined by, $$ T_0(\sum_{i=0}^px_i \mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i, \ \ T_0 = \text{Id}, $$ and for $i=1,\dots,p$, $$ T_i(\sum_{i=0}^p x_i\mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i + (x_p-ix_0)\mathbf{e}_i, $$ (these are all invertible except if $p=2$, where a shuffle as in Abhinav's answer works). The determinant is $F(x_0,x_1,\dots,x_p) = x_0x_p(x_p-1x_0)\cdots (x_p-(p-1)x_0)$, which vanishes identically on $\mathbb{F}_p^{p+1}$.

If $n \leq p$, then obviously this is true. For variables $x_1,\dots,x_n$, form the vector $\vec{v} = (x_1,\dots,x_n)^\dagger$, and consider the polynomial $F(x_1,\dots,x_n) = \text{det}(T_1\vec{v},\dots,T_n\vec{v})$. This is a homogeneous degree $n$ polynomial in the variables $x_1,\dots,x_n$ with coefficients in $\mathbb{F}_p$.

Now prove the following lemma: for every integer $m$, every homogeneous polynomial $F(x_1,\dots,x_m)$ of degree $\leq p$ that vanishes identically on $\mathbb{F}_p^m$ is the zero polynomial. This is proved by induction on $m$. For $m=1$, this is trivial. For $m=2$, this is a slight variant of the inhomogeneous analogue. Every inhomogeneous polynomial $F(x_1,1)$ that vanishes identically on $\mathbb{F}_p$ is a multiple of $x_1^p-x_1$. Since the degree is $\leq p$, this means that $F(x_1,x_2) = a(x_1^p-x_1x_2^{p-1})$ for some $a\in \mathbb{F}_p$. Since $F(1,0)$ is zero, also $a$ equals $0$, so that $F$ is the zero polynomial. That establishes the base case of the induction.

Assuming the result for $m-1$, write $F(x_1,\dots,x_m)$ as $$F(x_1,\dots,x_n) = F_n(x_1,\dots,x_{m-1})x_m^n + \dots + F_0(x_1,\dots,x_{m-1})x_m^0.$$ For fixed values $(a_1,\dots,a_{m-1})\in \mathbb{F}_p^{m-1}$ the resulting polynomial $F(a_1,\dots,a_{m-1},x) \in \mathbb{F}_p[x]$ vanishes identically on $\mathbb{F}_p$ by hypothesis. By the same argument as for $m=2$, $F(a_1,\dots,a_{m-1},x) = a(x^p-G_{p-1}(a_1,\dots,a_{m-1})x)$ with $G(a_1,\dots,a_{m-1}) = 1$. This forces $F_d(a_1,\dots,a_{m-1}) =0$ for $d\neq p,1$, which, by the induction hypothesis, forces the polynomials $F_d(x_1,\dots,x_{m-1})$ to also be zero. Finally plug in $(0,\dots,0,1)$ to conclude that the entire polynomial is zero.

Finally, for $n=p+1$, there is a generalization of Abhinav's counterexample. Namely, the $p+1$ matrices $T_0,\dots,T_p$ defined by, $$ T_0(\sum_{i=0}^px_i \mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i, \ \ T_0 = \text{Id}, $$ and for $i=1,\dots,p$, $$ T_i(\sum_{i=0}^p x_i\mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i + (x_p-ix_0)\mathbf{e}_i, $$ (these are all invertible except if $p=2$, where a shuffle as in Abhinav's answer works). The determinant is $F(x_0,x_1,\dots,x_p) = x_0x_p(x_p-1x_0)\cdots (x_p-(p-1)x_0)$, which vanishes identically on $\mathbb{F}_p^{p+1}$.

If $n \leq p$, then obviously this is true. For variables $x_1,\dots,x_n$, form the vector $\vec{v} = (x_1,\dots,x_n)^\dagger$, and consider the polynomial $F(x_1,\dots,x_n) = \text{det}(T_1\vec{v},\dots,T_n\vec{v})$. This is a homogeneous degree $n$ polynomial in the variables $x_1,\dots,x_n$ with coefficients in $\mathbb{F}_p$.

Now prove the following lemma: for every integer $m$, every homogeneous polynomial $F(x_0,\dots,x_m)$ of degree $\leq p$ that vanishes identically on $\mathbb{F}_p^m$ is the zero polynomial. This is proved by induction on $m$. For $m=0$, this is trivial. For $m=1$, this is a slight variant of the inhomogeneous analogue. Every inhomogeneous polynomial $F(1,x)$ that vanishes identically on $\mathbb{F}_p$ is a multiple of $x^p-x$. Since the degree is $\leq p$, this means that $F(x_0,x_1) = a(x_1^p-x_1x_0^{p-1})$ for some $a\in \mathbb{F}_p$. Since $F(0,1)$ is zero, also $a$ equals $0$, so that $F$ is the zero polynomial. That establishes the base case of the induction.

By way of induction, assume that $m\geq 2$, and assume the result for $m-1$. Write $F(x_0,\dots,x_m)$ as $$F_{0}(x_0,\dots,x_{m-1})x_m^n + \dots + F_{n-d}(x_0,\dots,x_{m-1})x_m^d + \dots + F_n(x_0,\dots,x_{m-1})x_m^0,$$
with each $F_{n-d}$ a homogeneous polynomial in $x_0,\dots,x_{m-1}$ of degree $n-d \leq p$. Plug in $(0,\dots,0,1)$ to conclude that the constant $F_0$ is zero. For fixed values $(a_0,\dots,a_{m-1})\in \mathbb{F}_p^{m}$ the resulting polynomial $F(a_0,\dots,a_{m-1},x) \in \mathbb{F}_p[x]$ vanishes identically on $\mathbb{F}_p$ by hypothesis. By the same argument as for $m=1$, $F(a_0,\dots,a_{m-1},x) = a(x^p-x)$ for some $a\in \mathbb{F}_p$. Since $F_0$ is zero, in fact $a$ equals $0$. Thus every $F_{m-d}(a_0,\dots,a_{m-1})$ equals $0$ for every $(a_0,\dots,a_{m-1})$ in $\mathbb{F}_p^m$. By the induction hypothesis, this forces every polynomial $F_{m-d}(x_0,\dots,x_{m-1})$ to be the zero polynomial. Therefore also $F(x_0,\dots,x_m)$ is the zero polynomial.

Finally, for $n=p+1$, there is a generalization of Abhinav's counterexample. Namely, the $p+1$ matrices $T_0,\dots,T_p$ defined by, $$ T_0(\sum_{i=0}^px_i \mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i, \ \ T_0 = \text{Id}, $$ and for $i=1,\dots,p$, $$ T_i(\sum_{i=0}^p x_i\mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i + (x_p-ix_0)\mathbf{e}_i, $$ (these are all invertible except if $p=2$, where a shuffle as in Abhinav's answer works). The determinant is $F(x_0,x_1,\dots,x_p) = x_0x_p(x_p-1x_0)\cdots (x_p-(p-1)x_0)$, which vanishes identically on $\mathbb{F}_p^{p+1}$.

If $p>n$, then obviously this is true. For variables $x_1,\dots,x_n$, form the vector $\vec{v} = (x_1,\dots,x_n)^\dagger$, and consider the polynomial $F(x_1,\dots,x_n) = \text{det}(T_1\vec{v},\dots,T_n\vec{v})$. This is a homogeneous degree $n$ polynomial in the variables $x_1,\dots,x_n$ with coefficients in $\mathbb{F}_p$.

Now prove the following lemma: for every integer $n<p$, every polynomial $F(x_1,\dots,x_m)$ of degree $\leq n$ that vanishes identically on $\mathbb{F}_p^m$ is the zero polynomial. This is proved by induction on $m$. For $m=1$, this is just the statement that a nonzero, degree $n$ polynomial can have at most $n$ zeroes. Assuming the result for $m-1$, write $F(x_1,\dots,x_m)$ as $$F(x_1,\dots,x_n) = F_n(x_1,\dots,x_{m-1})x_m^n + \dots + F_0(x_1,\dots,x_{m-1})x_m^0.$$ For fixed values $(a_1,\dots,a_{m-1})\in \mathbb{F}_p^{m-1}$ the resulting polynomial $F(a_1,\dots,a_{m-1},x) \in \mathbb{F}_p[x]$ vanishes identically on $\mathbb{F}_p$ by hypothesis. Thus, by the $m=1$ case above, the coefficients $F_d(a_1,\dots,a_{m-1})$ are all zero. Thus by the induction hypothesis, every coefficient polynomial $F_d(x_1,\dots,x_{m-1})$ is the zero polynomial. Therefore $F(x_1,\dots,x_m)$ is the zero polynomial.

If $n \leq p$, then obviously this is true. For variables $x_1,\dots,x_n$, form the vector $\vec{v} = (x_1,\dots,x_n)^\dagger$, and consider the polynomial $F(x_1,\dots,x_n) = \text{det}(T_1\vec{v},\dots,T_n\vec{v})$. This is a homogeneous degree $n$ polynomial in the variables $x_1,\dots,x_n$ with coefficients in $\mathbb{F}_p$.

Now prove the following lemma: for every integer $m$, every homogeneous polynomial $F(x_1,\dots,x_m)$ of degree $\leq p$ that vanishes identically on $\mathbb{F}_p^m$ is the zero polynomial. This is proved by induction on $m$. For $m=1$, this is trivial. For $m=2$, this is a slight variant of the inhomogeneous analogue. Every inhomogeneous polynomial $F(x_1,1)$ that vanishes identically on $\mathbb{F}_p$ is a multiple of $x_1^p-x_1$. Since the degree is $\leq p$, this means that $F(x_1,x_2) = a(x_1^p-x_1x_2^{p-1})$ for some $a\in \mathbb{F}_p$. Since $F(1,0)$ is zero, also $a$ equals $0$, so that $F$ is the zero polynomial. That establishes the base case of the induction.

Assuming the result for $m-1$, write $F(x_1,\dots,x_m)$ as $$F(x_1,\dots,x_n) = F_n(x_1,\dots,x_{m-1})x_m^n + \dots + F_0(x_1,\dots,x_{m-1})x_m^0.$$ For fixed values $(a_1,\dots,a_{m-1})\in \mathbb{F}_p^{m-1}$ the resulting polynomial $F(a_1,\dots,a_{m-1},x) \in \mathbb{F}_p[x]$ vanishes identically on $\mathbb{F}_p$ by hypothesis. By the same argument as for $m=2$, $F(a_1,\dots,a_{m-1},x) = a(x^p-G_{p-1}(a_1,\dots,a_{m-1})x)$ with $G(a_1,\dots,a_{m-1}) = 1$. This forces $F_d(a_1,\dots,a_{m-1}) =0$ for $d\neq p,1$, which, by the induction hypothesis, forces the polynomials $F_d(x_1,\dots,x_{m-1})$ to also be zero. Finally plug in $(0,\dots,0,1)$ to conclude that the entire polynomial is zero.

Finally, for $n=p+1$, there is a generalization of Abhinav's counterexample. Namely, the $p+1$ matrices $T_0,\dots,T_p$ defined by, $$ T_0(\sum_{i=0}^px_i \mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i, \ \ T_0 = \text{Id}, $$ and for $i=1,\dots,p$, $$ T_i(\sum_{i=0}^p x_i\mathbf{e}_i) = \sum_{i=0}^p x_i \mathbf{e}_i + (x_p-ix_0)\mathbf{e}_i, $$ (these are all invertible except if $p=2$, where a shuffle as in Abhinav's answer works). The determinant is $F(x_0,x_1,\dots,x_p) = x_0x_p(x_p-1x_0)\cdots (x_p-(p-1)x_0)$, which vanishes identically on $\mathbb{F}_p^{p+1}$.