Vector spaces span

I didnt understand when you said to eliminate x4 in each equation because I ended up with three identical equations.

I followed this method on this video
https://www.youtube.com/watch?v=GgSOb0bnfjE&feature=youtube_gdata_player

But what would I write on my question marks?

You're expecting to end up with identical equations. That's what it means for a system of equations to have a kernel.


What allowable x4 are there such that (reduced matrix).(x1,x2,x3,x4) has a chance of being (0,0,0)? What, if anything, does that force the other variables to obey? Then what allowable x3 are there, satisfying those forcing constraints, etc? Repeat until you have constraints on all of x1,x2,x3,x4. (I don't know if your reduced form is right - I didn't bother checking that.)

Hi again. Could yoy tell me how to find basis of a row space of a matrix please?

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Hi again. Could yoy tell me how to find basis of a row space of a matrix please?

Call your vectors U, V, W. First you know that <U,V,W> must have dimension 1,2 or 3 (since it is spanned by a set of size 3). You should be able to work out quickly that it cannot have dimension 1. Now try to solve aU+bV+cW=0 and one of two things will happen.

1. You find that a=b=c=0 is the only solution, in which case the vectors must be independent so they already form a basis (linearly independent and a spanning set).

2. You get a non-zero solution for a,b,c and so the vectors are not linearly independent. Since you know it does not have dimension 1, any 2 of the 3 vectors will form a basis (can you see why?).

If theyre not LI, then how did you know that it would br any of tye two instead of any if the one?


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What would it mean if any one of the row vectors formed a basis of the row space?

What would it mean if any one of the row vectors formed a basis of the row space?

I did it the equation way too. It seems like that the basis of a row space will always just be the same as the row space.
I think that I still dont understand tbh.
Definition of a basis is that it has to span the vectors in the set and be LI. The set of rowA is LI so it must be a basis if rowA too. ?

Consider the matrix {{1,0}, {1,0}, {1,2}} which has three rows. Find a basis of its row space.

If theyre not LI, then how did you know that it would br any of tye two instead of any if the one?


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Ignore first pic.
? If it is then ok I see that not all of them are in. And yes I do see that you've repeated a vector so the set is not LI. Im just getting used to this matrix method.

That wouldnt be classed as a set wpuld it? Cuz it has a repeated element. Im guessing that you were just showing a point.

Which exam board is this? Is this fp3???

Ignore first pic.
? If it is then ok I see that not all of them are in. And yes I do see that you've repeated a vector so the set is not LI. Im just getting used to this matrix method.

That wouldnt be classed as a set wpuld it? Cuz it has a repeated element. Im guessing that you were just showing a point.

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I did it the equation way too. It seems like that the basis of a row space will always just be the same as the row space.

I think that I still dont understand tbh.
Definition of a basis is that it has to span the vectors in the set and be LI. The set of rowA is LI so it must be a basis if rowA too. ?

This is definitely not true. In fact, there is no situation in which they can be the same. The basis of any vector space is always a strict subset of that vector space (in particular, it doesn't include the zero of that vector space).

He's using mathematica-style matrix formatting. He's asking you to find a basis of the row space of the matrix

$\displaystyle\left( \begin{array}{cc}1 & 0 \\ 1 & 0 \\ 1 & 2\end{array}\right)$




This is definitely not true. In fact, there is no situation in which they can be the same. The basis of any vector space is always a strict subset of that vector space (in particular, it doesn't include the zero of that vector space).



You are confusing the set of rows of $A$ with its row space. The former is the finite set containing its rows expressed as elements of $\mathbb{R}^n$ (where $A$ has $n$ columns). The latter is the (infinite, unless you're dealing with a finite field or a zero matrix) vector space spanned by them.

Specifically, if we let $R$ denote the row space of an $n$ by $m$ matrix (where $v_1, v_2, ..., v_m \in \mathbb{R}^n$ are the rows of $A$), then
$\displaystyle R = \left\{ \sum_{k=1}^m \alpha_k v_k : \alpha_k \in \mathbb{R} \right\}$
(assuming real matrices throughout. Obviously, it works fine for matrices over any ring).


It is, however, true that if the rows of a matrix are linearly independent, then they must form a basis of the row space of that matrix, since they span it by definition.

.

I think he meant "the rows always form a basis", which is counterexampled by my repeated-row matrix.