Representing vectors in rn using subspace members | Linear Algebra | Khan Academy

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Showing that any member of Rn can be represented as a unique sum of a vector in subspace V and a vector in the orthogonal complement of V.
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[@kartikkamboj295]

Absolutely amazing! Hats off. 👏🏻

[@OsamaComm]

I think we need to condition at 05:40 that (V U V_OrthognalComplement = R^n) so we can safely say dim(V) + dim(V_complement) = n.

[@SomethingSoOriginal]

At 22:39 does that mean that Rn = span(V) + span(Vperp) ? Meaning V and its orthogonal cover all elements of Rn?

[@Yh-gu3cw]

good explanation.

[@turningheadfart]

nice tut.

[@khamilkhan4285]

GOAT

[@debendragurung3033]

26:52; theres a uique way to express a vector as a combination of two orthogonal vectors besides zero vector...

[@tw0ey3dm4n]

it took a while to get it (i think)... you let the linear combination of V vectors + the linear combinations of orth(V) vectors = 0 vector. then through some slight rearranging by taking the orth(V) vectors to the right hand side, we see that any vector x in V, also belongs to orth(V) means that x dot x = 0 ==> x = 0 vector is the only solution... this implies the linear combination of V vectors can only be 0, and the linear combination of orth(V) vectors can only be 0 means the linear combination of both V and orth(V) vectors can only be 0 if all the scalars are 0 ==> vectors of V + vectors of orth(V) are linearly independent and form a basis to span Rn. i had to type this down to see if i got this right?

[@Jeff-wc5ho]

In our initial example, f the dimension of V doesn't equal the dimension of Vperp, then how can x be an element of both? That doesn't make any sense.

[@Cashman9111]

yeah, he could mention it's only for k=n/2, otherwise it's very trivial

[@Cashman9111]

still, zero vector in R3 is not the same as zero vector in R2, so I'm not quite sure

Sorry for being so late but this is an important question so it deserves an answer. I think you may be getting a bit confused with the way vectors are written. If V is a subspace of dimension, say 2, that doesn't mean that if x is a member of V, then x must be something like x = (x1, x2). Take for instance the example of a plane in R^3 that passes through the origin. That is a subspace of IR^3 which has dimension 2 but it's members are written as x = (x1, x2, x3). In fact, IR^3 is by definition a subspace of itself and it has to contain this plane. There you have the intersection of two subspaces with different dimensions.

[@TomCurry21]

this stuff is wack