Subspaces

Subspaces of a vector space. Sums and direct sums.

[thomasbates]

Thank you for this video! I was reading your book and came across "The set of differentiable real-valued functions f on the interval (0, 3) such that f`(2) = b is a subspace of R^(0, 3) if and only if b = 0" and I at first did not understand why this is true. I then went to this video and because of the precise language you used to explain it, I was able to figure out why the assertion is in fact true! It did take me about 30 minutes but it was so worth it! Thank you for including that assertion in the examples.

[akoh]

Thank you so much for making these. I'm re-learning the subject and these videos are a great supplement to the book.

[omarburhani]

Thank you so much!
students from Syria are benefiting

[PC.]

Unbelievable its going dam easy for me now...please do this kind of videos on another subjects..please

[stefanoplaino]

Example 4. Is it right to proof that b be equal to 0 as follows: let f(x) and g(x) be elements of our set. As such by additivity (f+g)(x) = f(x) + g(x) is also element of our set. That implies that (f+g)'(x) = b = f'(x) + g'(x) = b + b = 2b. That holds if and only if b = 0. Sounds correct? Thank you for any comment or reply.

[TheMrDJD]

3:31 As you build up the set for the subspace of F^4 it's not clear why b has to be 0 for this to be a subspace of F^4 for two reasons (though I'm probably not seeing something obvious here).

1. Couldn't the zero vector just be defined as (0, 0, -b, 0) when x_3=5x_4+b or is it because F^4 is implicitly (0, 0, 0, 0) and therefore F^4 and this subspace would have different zero vectors?

2. Similar to 1, instead of the 0 vector accounting for the b, x_4 in F^4 could be defined to be [ x_4-(1/5)* b ] (one depth of recursion so this doesn't become an infinite loop). Therefore x_3 then becomes:

x_3 = 5x_4+b = 5(x_4-(1/5)* b) + b = 5x_4

If this is the definition of x_4, the b wouldn't need to be reflected in the zero vector as (0, 0, 0, (1/5)*b) and would still be (0, 0, 0, 0).

In which case scenario #2 would be a subspace of F^4, and b wouldn't have to equal 0, right?

[michelef]

Around 3:33, can i show that b must be = 0 for us to have a subspace in a slightly different way? Namely, if i take a list (y1, y2, 5y4 + b, y4) and add it to the given list i get that the fourth element is 5y4 + 5x4 + 2b. For the subset to be close under addition 2b must be equal to b, which is only true for b=0.

[twisthz]

Dear Prof. Axler, I am having difficulty getting my head around the example for Sum of subspaces, that is, Example 1.38, pg.20 of your textbook. Specifically, when you define the subspace U+W, why did you set the third element to be "y" and the fourth element to be "z" ?

[onlyumichaellee]

I have a question about the example 1.35(a), I understood that in the set U, the additive identity is (0, 0, 0, 0), and (0, X2, X3, X4) is a subspace of F^4. But why does b has to 0, in the e.x.1.35, and only has to 0 to satisfies it is a subspace of F^4, though I notice the subspace definition implies that (X1, X2, X3, 0) is a subspace of F^4, but I don't understand the relationship of b with other Xs.

[Cpt.Zenobia]

in example (4) @5:09 does f'(x) = 0, where x in (0, 3), means that the only functions in R^(0, 3) are all the constant functions f(x) = C?

[jshook]

can someone clarify why 4th example of a subspace at 5:28 is in fact a subspace? If the set f consists of differentiable functions, and not differentiated functions, then why does the fact that a differentiated function = 0 matter? is there something obvious that I am missing? thanks.

[oluwaseuncardoso]

I am a bit confused with in 4:11 . How is it clear that 0 is in the set of continuous real-valued functions on the interval[0, 1]? All we know is that the function is continuos, but we don't know the range it maps to. The range it maps to may not include 0.

[laharalal]

When considering the additive identity in the sum of subspaces of vector space V, there is additive identity, although it is not unique (consider W+W and W+{0} ) but ¿what about additive inverse? you can write 0=v+w as v in W and w=-v in W so 0 is in W+W and as many W+X as X's contains W. But there are infinitely many vectors in W+X, there is no way that there exists subspaces X, Y such that X+Y={0} (but i don't know how to prove it). On the other hand, if X, Y are not subspaces, you possibly have X+Y={0}, but this is not the case i'm interested.

Edit: as i finished writing this comment i realized that the only subspace with additive inverse is zero,
{0} + {0} = {0}. Still don't know how to prove that if X, Y are not {0} then X+Y is not {0}. So another question I am thinking that if a vector space is not zero then it has infinite vectors, but what happens when the field is finite, can i have a vector space of finite elementes different from the trivial?

[OneirosProduction]

You don't have to check that *0* is an element of your subset to prove it's a subspace, you only need to check that the subset is not empty.

Indeed, if v is an element of the subset and you have proven that it is closed under linear combinations, then *0* = 0v must also be an element of that subspace.

[sarahmedouni]

what does "sequence of complex numbers with limit 0" mean?

[glauconariston]

Why aren't sets of lines through the origin in R^2 also subspaces of R^2? They seem to hold up to all the properties as far as I can see. Anyone know a counterexample?

[mrnarason]

Another example, R^3 is the direct sum of the z-axis and the xy-plane.

However, R^3 is NOT the direct sum of the zy-plane and the xy plane, because the decomposition is not unique.

[DonKarelove]

I find the analogy of the sum of subspaces and the union of sets rather confusing.
The sum implies that in the resulting subspace contains not only those vectors that were in the original subspaces, but also all their sums. Why make this comparison if this important property doesn't hold for unions?

[alejandrodeharo]

A function f W R ! R is called periodic if there exists a positive number
p such that f .x/ D f .x C p/ for all x 2 R. Is the set of periodic
functions from R to R a subspace of RR? Explain. help