Linear Algebra II

September 15, 2014


Filed under: 2014 Fall — Y.K. Lau @ 8:28 PM

Today we further discuss Remark 1 in the last post – what happens when we perform set operations on two subspaces. A key point is that we are led to the concepts of the sum of two subspaces, and the direct sum. (The lecture slides are uploaded in the folder Slides at moodle.)

After the lecture ended, some classmates asked questions that I would share with you here.

  • Qn 1. Is the direct sum (of subspaces) related to quotient vector spaces?

    You may wonder why your classmate ask this question because they (direct sum and quotient space) seem to be very different objects. If the vector space {V=U\oplus W}, then {U} and {W} are containing vectors in {V}. However, {V/W} contains the equivalence classes {[v]} where {v\in V}, which is more abstract and complicated. Nevertheless, there are interesting relations between the concepts of direct sum and quotient vector space. You will get some ideas in Tutorial 2 and Assignment 2.

  • We explained in lecture that even if {X} and {Y} are subspaces of {V}, {X\setminus Y} is NOT a subspace of {V}. Our proof is that both {X} and {Y} contain the zero vector {\underline{0}}; thus {\underline{0}\notin X\setminus Y} and so {X\setminus Y} is not a subspace.

    Qn 2. A classmate asked why the zero vector of {X} must equal the zero vector of {Y}.

    (Well our proof does not work if they are different.) This question sounds silly but indeed it makes sense very much! If you were me, how would you reply?

    Mine is: the zero vector of any subspace (of {V}) has to be the zero vector of {V}.

    First of all, zero vector in a vector space is unique (i.e. given a vector space, there is only one element fulfilling (A4)). [Prove it! If you cannot work it out, please ask me or our tutor.]

    Next, let {V} be a vector space and {X} be its subspace. Suppose {0_V} is the zero vector of {V} and {0_X} is the zero vector of {X}. Now {0_X\in X\subset V}, i.e. {0_X} is a vector in {V}. Write {0} for the real number zero and use Part (1) of Theorem 6.1.3 (i.e. Thm 3 of Section 6.1) twice, we infer

    \displaystyle  0_V\stackrel{{\rm Thm \,6.1.3 (1)\, for} \, V}{=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!} \displaystyle  \ \ 0 \displaystyle\cdot 0_X \stackrel{{\rm Thm \,6.1.3 (1)\, for} \, X}{=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!=\!\!\!}\ \ 0_X.





  1. I guess there is a typo in the ‘mine is’ line.

    Comment by Shum Ho Pan — September 16, 2014 @ 12:58 AM | Reply

    • Thanks. Amended.

      Comment by Anonymous — September 16, 2014 @ 10:23 AM | Reply

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at

%d bloggers like this: