Dim null ST <= dim null S + dim null T

  • #1
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Homework Statement:

Suppose ##U## and ##V## are finite dimensional vector spaces and ##S \in \mathcal{L}(V, W)## and ##T \in \mathcal{L}(U, V)##. Prove that ##\dim \operatorname{null} ST \le \dim \operatorname{null} S + \dim \operatorname{null} T##.

Relevant Equations:

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For a linear map ##T \in \mathcal{L}(U, V)##, ##\text{null} T = \lbrace u \in U : Tu = 0 \rbrace##

Fundamental theorem of linear maps: ##\dim U = \dim \text{null} T + \dim T(U)##
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Proof: Since ##U, V## are finite dimensional, we have that ##\operatorname{null} S, \operatorname{null} T## are finite dimensional. Let ##v_1, \dots v_m## be a basis of ##\operatorname{null}S## and ##u_1, \dots, u_n## be a basis of ##\operatorname{null} T##. It is enough to show there are ##m + n## vectors that span ##\operatorname{null} ST##. Let ##x \in \operatorname{null}ST##. Then ##x \in \operatorname{null}T## or ##Tx \in \operatorname{null}S##. So ##u_1, \dots, u_n## is in our list of ##m+n## vectors.

If ##T## maps onto ##\operatorname{null}S##, then for each ##v_i## there is ##y_i \in U## such that ##Ty_i = v_i##. So ##Ty_1, \dots, Ty_m## spans ##\operatorname{null}S##. This implies ##\operatorname{span} \lbrace y_1, \dots, y_m, u_1, \dots, u_n \rbrace\supseteq \operatorname{null}ST##. But I don't think I can say ##T## maps onto ##\operatorname{null}S##. Can I have a hint, please?
 

Answers and Replies

  • #2
Infrared
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The approach of finding a spanning set of at most ##m+n## should work, and the ##u_i## should be included. Instead of taking these vectors ##y_i##, how about taking a basis for ##\text{Range}(T)\cap\text{Nul}(S)##?

Edit: To clarify, if ##Tx_1,\ldots, Tx_n## is such a basis, I mean to use the vectors ##x_1,\ldots, x_n.##
 
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  • #3
Math_QED
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Hint: ##\operatorname{null}(ST) = T^{-1}(\operatorname{null}(S))## and rank-nullity theorem.
 
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  • #4
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Thank you, I think I got it using what you said:

Proof: Since ##U, V## are finite dimensional, we have ##\operatorname{null} S## and ##\operatorname{null}T## are finite dimensional. Let ##u_1, \dots, u_n## be a basis for ##\operatorname{null} T## and ##Tx_1, \dots, Tx_k## be a basis for ##\operatorname{Range}(T) \cap \operatorname{null} S##. We note that ##k \le \dim \operatorname{null} S##. We'll show the statement ##\operatorname{span} \lbrace u_1, \dots, u_n, x_1, \dots, x_k \rbrace = \operatorname{null} ST## is true.

##(\subseteq)## Let ##a_1u_1 + \dots a_nu_n + b_1x_1 + \dots + b_kx_k \in \operatorname{span} \lbrace u_1, \dots, u_n, x_1, \dots, x_k \rbrace##. Then

##
\begin{align*}
ST(a_1u_1 + \dots a_nu_n + b_1x_1 + \dots + b_kx_k) &= S(a_1T(u_1) + \dots + a_n T(u_n) + b_1T(x_1) + \dots + b_kT(x_k)) \\
& = S(0 + \dots + 0 + b_1T(x_1) + \dots + b_kT(x_k)) \\
&= S(b_1T(x_1) + \dots + b_kT(x_k)) \\
&= 0
\end{align*}##
This shows ##\operatorname{span} \lbrace u_1, \dots, u_n, x_1, \dots, x_k \rbrace \subseteq \operatorname{null} ST##.

##(\supseteq)## If ##x \in \operatorname{null}ST##, then ##x \in \operatorname{null}T## or ##Tx \in \operatorname{null} S##. I.e., ##x \in \operatorname{span} \lbrace u_1, \dots, u_n \rbrace##. Otherwise, ##Tx \in \operatorname{Range}(T) \cap \operatorname{null}S## and since ##x_1, \dots, x_k## is linearly independent, we have ##x \in \operatorname{span} \lbrace x_1, \dots, x_k \rbrace##. This shows ##\operatorname{span} \lbrace u_1, \dots, u_n, x_1, \dots, x_k \rbrace \supseteq \operatorname{null} ST##.

We can conclude ##\dim \operatorname{null}ST \le n + k \le n + m = \dim \operatorname{null}T + \dim \operatorname{null}S##. []
 
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  • #5
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I see using post #3:

Proof: First we show ##\operatorname{null}(ST) = T^{-1}(\operatorname{null}(S))##.

##(\subseteq)## Let ##x \in \operatorname{null}(ST)##. Then ##S(Tx) = 0## which means ##Tx \in \operatorname{null}(S)## i.e. ##x \in T^{-1}(\operatorname{null}(S))##.

##(\supseteq)## Let ##x \in T^{-1}(\operatorname{null}(S))##. Then ##Tx \in \operatorname{null}(S)## so ##S(Tx) = 0## i.e. ##x \in \operatorname{null}(ST)##.

We can conclude ##\operatorname{null}(ST) = T^{-1}(\operatorname{null}(S))##.

By rank nullity theorem: ##\dim \operatorname{null}(ST) = \dim T^{-1}(\operatorname{null}(S)) = \dim \operatorname{null}(T) + \dim T(T^{-1}(\operatorname{null}(S))) \le \operatorname{null}(T) + \operatorname{null}(S)##. []
 
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  • #6
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Or you can consider that ##KerT \subset KerST##. How much larger can it be?
 

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