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

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In summary, the conversation discusses the goal of showing that def(SoT) <= def(T)+def(S). It is mentioned that Ker(SoT) is not a subset of Ker(S)+Ker(T) and that the problem can be solved by starting with Ker(T). The conversation also suggests choosing a basis for W and using the set {u1,...,ur,v1,...,vk} to span Ker(SoT).
  • #1
Denis99
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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.
 
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  • #2
Denis99 said:
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.
 
  • #3
Opalg said:
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 :)
 

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

What does the expression "dim(Ker(SoT))" mean?

The expression "dim(Ker(SoT))" refers to the dimension of the kernel of the composition of two linear transformations, S and T.

What is the significance of the inequality "dim(Ker(SoT))<=dim(KerT)+dim(KerS)"?

The inequality "dim(Ker(SoT))<=dim(KerT)+dim(KerS)" is a fundamental property of linear transformations that relates the dimensions of the kernels of two linear transformations and their composition.

How is the inequality "dim(Ker(SoT))<=dim(KerT)+dim(KerS)" derived?

The inequality "dim(Ker(SoT))<=dim(KerT)+dim(KerS)" can be derived using basic properties of linear transformations, such as the fact that the dimension of the kernel of a linear transformation is equal to the difference between the dimension of the domain and the dimension of the range.

What is the practical application of the inequality "dim(Ker(SoT))<=dim(KerT)+dim(KerS)"?

The inequality "dim(Ker(SoT))<=dim(KerT)+dim(KerS)" is useful in various areas of mathematics and science, such as linear algebra, differential equations, and control theory. It allows us to determine the maximum number of linearly independent solutions of a system of linear equations.

Are there any exceptions to the inequality "dim(Ker(SoT))<=dim(KerT)+dim(KerS)"?

Yes, there are exceptions to the inequality "dim(Ker(SoT))<=dim(KerT)+dim(KerS)" when the dimensions of the kernels of the individual linear transformations are not independent. In such cases, the inequality may not hold true.

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