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

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In summary, the proof shows that there is a spanning set of at most ##m+n## vectors that span ##\operatorname{null} ST##. The proof also shows that if ##Tx_1,\ldots, Tx_k## is such a spanning set, then for each ##v_i## there is ##y_i \in U## such that ##Ty_i = v_i##.
  • #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?
 
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  • #2
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
Hint: ##\operatorname{null}(ST) = T^{-1}(\operatorname{null}(S))## and rank-nullity theorem.
 
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  • #4
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
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
Or you can consider that ##KerT \subset KerST##. How much larger can it be?
 

1. What does the equation "Dim null ST <= dim null S + dim null T" mean?

The equation represents the dimension of the null space of the combined matrix ST, which is less than or equal to the sum of the dimensions of the individual null spaces of matrices S and T.

2. How is the equation "Dim null ST <= dim null S + dim null T" useful in scientific research?

This equation is useful in linear algebra and matrix theory, particularly in understanding the properties of null spaces and their relationship to linear transformations.

3. Can you provide an example to illustrate the equation "Dim null ST <= dim null S + dim null T"?

Say we have two matrices, S with a null space of dimension 3 and T with a null space of dimension 2. The combined matrix ST will have a null space of dimension 5, which is equal to the sum of the dimensions of the individual null spaces (3+2).

4. Is the equation "Dim null ST <= dim null S + dim null T" always true?

Yes, the equation is always true. This is because the null space of a combined matrix cannot have a higher dimension than the sum of the dimensions of the individual null spaces.

5. What other important properties or relationships can be derived from the equation "Dim null ST <= dim null S + dim null T"?

The equation can also be used to prove that the null space of a matrix product is a subset of the intersection of the null spaces of the individual matrices. It also helps in understanding the concept of linear independence and dependence in vector spaces.

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