number of bent components, then its algebraic degree is at most $k$.

For any non-bent component $f$ of $F$, let $g$ be a bent component, then by Theorem \[number\] $f+g$ is bent. One then immediately gets the desired assertion by the upper bound for the algebraic degree of bent functions.

For any vectorial bent $(n,k)$-function $F(x)$ ($n=2k$), with the inclusion map $i: \mathbb{F}^{k}_{2}\rightarrow \mathbb{F}^{m}_{2}$, $a\mapsto (a,0)$, it can also be regarded as an $(n,m)$-function which can be easily seen to have the maximal number $2^{m}-2^{m-k}$ of bent components. For this reason, in this paper we call such $(n,m)$-function $G(x)=(F(x),0)$ trivial.

As it has been pointed out by Pott et al., for an $(n,n)$-function $F$ with maximal number of bent components, it can produce many vectorial bent $(n,k)$-functions [@Pott Proposition 3]. Explicitly, if ${\rm Tr}^{n}_{1}(\alpha F(x))$ is bent for any $\alpha \in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$, then ${\rm Tr}^{n}_{k}(\alpha F(x))$ is a vectorial bent function for any $\alpha \in \mathbb{F}_{2^{n}}\backslash \mathbb{F}_{2^{k}}$. With this observation, we revisit the following $(n,n)$-function $F(x)$ which is essentially due to Leander and Kholosha [@Leander-Kholosha].

[**Example 1.**]{} Let $n=2k$, $F(x)=\sum\limits^{2^r-1}_{i=1}x^{(i2^{k-r}+1)(2^k-1)+1}$ with $1 \leq r\leq k$, and ${\rm gcd}(r,k)=1$. It has been observed by Li et al. that, for any 