[proofplan] The proof uses the scaling property of homogeneous polynomials. Since $f$ is homogeneous of degree $d$, evaluating $f$ at a scalar multiple of a vector multiplies the value by $\lambda^d$. Applying this identity to the zero $a$ gives $f(\lambda a)=\lambda^d f(a)=0$ for every scalar $\lambda \in k$. [/proofplan]

[step:Apply the homogeneous scaling identity to the point $a$]Let $F: k^n \to k$ denote the polynomial function induced by $f$, so $F(b)=f(b)$ for every $b \in k^n$. Since $f$ is homogeneous of degree $d$, the scaling characterisation of homogeneous polynomials applies. [citetheorem:9252] Its hypotheses are satisfied by the given field $k$, integer $n$, polynomial $f$, and degree $d$. Hence, for every $\lambda \in k$ and every $b \in k^n$, \begin{align*} F(\lambda b)=\lambda^d F(b). \end{align*} Taking $b=a$ gives \begin{align*} f(\lambda a)=\lambda^d f(a). \end{align*}[/step]

[guided]Let $F: k^n \to k$ be the polynomial function represented by $f$, meaning that $F(b)=f(b)$ for each vector $b \in k^n$. The point of introducing $F$ is only to make explicit that we are evaluating the polynomial on vectors in $k^n$. The hypothesis says that $f$ is homogeneous of degree $d$. Therefore the [scaling characterization of homogeneous polynomials](/theorems/9252) applies. [citetheorem:9252] Its hypotheses are exactly that $k$ is a field, $n \in \mathbb{N}$, $d \in \mathbb{N} \cup \{0\}$, and $f \in k[x_1,\ldots,x_n]$ is homogeneous of degree $d$. The conclusion is that scaling every coordinate by $\lambda$ scales the value of the polynomial by $\lambda^d$. Thus, for every $\lambda \in k$ and every $b \in k^n$, \begin{align*} F(\lambda b)=\lambda^d F(b). \end{align*} Now substitute the particular vector $a \in k^n$ into this identity. This gives \begin{align*} f(\lambda a)=\lambda^d f(a). \end{align*}[/guided]

[step:Use that $a$ is a zero of $f$] The hypothesis gives $f(a)=0$. Therefore, for every $\lambda \in k$, \begin{align*} f(\lambda a)=\lambda^d f(a)=\lambda^d \cdot 0=0. \end{align*} Thus every scalar multiple $\lambda a$ is also a zero of $f$, as required. [/step]