Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Dim null ST <= dim null S + dim null T
Reply to thread
Message
[QUOTE="fishturtle1, post: 6369933, member: 606256"] 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##. [] [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Dim null ST <= dim null S + dim null T
Back
Top