Dimension of a set with vector function

[fcoulomb]

I have a vector $\textbf{v} \in \mathbb{R}^{3N}$ and a function $\textbf{Ψ} : \mathbb{R}^{3N} \longrightarrow \mathbb{R}^p$
such that $\textbf{Ψ}(\textbf{v})=0$.

Why the set $T=\{ \textbf{x} \in \mathbb{R}^{3N} \ | \ \textbf{Ψ}(\textbf{x})=0 \}$ has dimension $n=3N-p$?

 

[member 587159]

I have a vector $\textbf{v} \in \mathbb{R}^{3N}$ and a function $\textbf{Ψ} : \mathbb{R}^{3N} \longrightarrow \mathbb{R}^p$
such that $\textbf{Ψ}(\textbf{v})=0$.

Why the set $T=\{ \textbf{x} \in \mathbb{R}^{3N} \ | \ \textbf{Ψ}(\textbf{x})=0 \}$ has dimension $n=3N-p$?

There are surely details missing. Assuming that your function $\Psi$ is linear, we have that $T = ker \Psi$ and $\dim( ker \Psi) = 3N - p$ if and only if $\Psi$ is surjective where we used the rank-nullity theorem (also known as second dimension theorem).