# Some infinite dimensional linear algebra, kernels of linear maps

I'm studying functional analysis (namely weak convergence) and need to prove the following result: if $f,f_1,\ldots f_n$ are some linear maps $X\to \mathbb{C}$, where $X$ is a vector space over $\mathbb{C}$ then the inclusion $\bigcap\mathrm{Ker}f_i\subset \mathrm{Ker}f$ implies that $f\in \mathrm{Span}(f_1,\ldots f_n)$. It can be easily seen in the case $n=1$: $\mathrm{Ker}f_1\subset \mathrm{Ker}f$ implies that $\mathrm{Ker}f_1=\mathrm{Ker}f$ and immediately $f_1=cf$ for some $c\in \mathbb{C}$. Could you help me with the general case?

• can you please explain why $Kerf1⊂Kerf$implies that $Kerf1=Kerf$ Jun 1, 2014 at 9:52
• Use strong induction. For the inductive step, apply the inductive hypothesis to $f$, $f_1$, $\ldots$, $f_n$ restricted to $\text{ker}(f_{n+1})$. Jun 1, 2014 at 9:56
• Your assertion "Ker$f_1\subset$Ker$f \Rightarrow$Ker$f_1=$Ker$f$" seems false. Consider $f_1:\mathbb C[X] \to \mathbb C, 1 \mapsto 0, X^i \mapsto X^i$ for $i>0$ and $f:\mathbb C[X] \to \mathbb C, 1 \mapsto 0, X\mapsto 0, X^i \mapsto X^i$ for $i>1$. Jun 1, 2014 at 9:56
• For finite-dimensional $X$ and $n=1$ the statement in the question is certainly true. One has just to omit the claim $\ker f_1 = \ker f$ and put $f = cf_1$ instead of $f_1 = cf$ (allowing for $c\in C$ to be $0$). Jun 1, 2014 at 10:09
• That assertion of the OP is correct, I think... Jun 1, 2014 at 10:10

The map $(f_1,\dots,f_n):V\to\mathbb C^n$ factors through the injective map $F:V/\bigcap\mathrm{Ker}f_i\to\mathbb C^n$. Since $\bigcap\mathrm{Ker}f_i\subset \mathrm{Ker}f$, the map $f:V\to\mathbb C$ factors through $\tilde f:V/\bigcap\mathrm{Ker}f_i\to\mathbb C$. Since $F$ is injective, we can extend the linear form $\tilde f$ to a linear form on $\mathbb C^n$ (i.e. $\tilde f$ is the composition $V/\bigcap\mathrm{Ker}f_i\to\mathbb C^n\to\mathbb C$). If that extended form sends the $i$-th basis vector of $\mathbb C^n$ to $c_i$ then $f=\sum_i c_i f_i$.