Show that dim(Ker(SoT))<=dim(KerT)+dim(KerS)

[Denis99]

Like in the topic, the goal is to show that def(SoT) <= def(T)+def(S) (where def(P)=dim(KerP), T,S:V -> V are linear transformations and V<infinity).
Unfortunately Ker(SoT) isn`t a subset of Ker(S)+Ker(T), so I try to solve this problem starting with that Ker(T) is subset of Ker(SoT), but I don`t know if this is a good idea.

 

[Opalg]

Like in the topic, the goal is to show that def(SoT) <= def(T)+def(S) (where def(P)=dim(KerP), T,S:V -> V are linear transformations and V<infinity).
Unfortunately Ker(SoT) isn`t a subset of Ker(S)+Ker(T), so I try to solve this problem starting with that Ker(T) is subset of Ker(SoT), but I don`t know if this is a good idea.

Hi Denis99, and welcome to MHB.

Yes, definitely start with $\ker(T)$. You need to know how much of the range of $S$ lies in $\ker(T)$. So let $\def\ran{\operatorname{ran}}W = \ran(S) \cap\ker(T)$, and let $k = \dim(W)$. (Notice that $k\leqslant \dim(\ker T)$.) Choose a basis $\{w_1,\ldots,w_k\}$ for $W$. For $1\leqslant i\leqslant k$, choose $v_i\in V$ with $Sv_i = w_i$ (this is possible because each $w_i$ is in the range of $S$).

Now let $\{u_1,\ldots,u_r\}$ be a basis for $\ker(S)$, where $r = \dim(\ker S)$. Show that the set $\{u_1,\ldots,u_r,v_1,\ldots,v_k\}$ spans $\ker(S\circ T)$ in order to get your result.

 

[Denis99]

Hi Denis99, and welcome to MHB.

Yes, definitely start with $\ker(T)$. You need to know how much of the range of $S$ lies in $\ker(T)$. So let $\def\ran{\operatorname{ran}}W = \ran(S) \cap\ker(T)$, and let $k = \dim(W)$. (Notice that $k\leqslant \dim(\ker T)$.) Choose a basis $\{w_1,\ldots,w_k\}$ for $W$. For $1\leqslant i\leqslant k$, choose $v_i\in V$ with $Sv_i = w_i$ (this is possible because each $w_i$ is in the range of $S$).

Now let $\{u_1,\ldots,u_r\}$ be a basis for $\ker(S)$, where $r = \dim(\ker S)$. Show that the set $\{u_1,\ldots,u_r,v_1,\ldots,v_k\}$ spans $\ker(S\circ T)$ in order to get your result.

Thank you very much! That solution is brilliant :)