Linear algebra, basic vector space proofs

A

AbdelRahmanShady

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I am trying to read a book about vector spaces, without taking introductory courses about linear algebra to be rigorous from start. Any way I stumbled across
this proof

1647983053385.png

and my problem is in proof b.
1647983123625.png

cant he use vs 6. what prevents him from doing so.
then (-a)x = (-1 * a)x = -(ax) = (a * -1)x = a(-x)
what is error in my proof so he didnt use it
 
AbdelRahmanShady said:
I am trying to read a book about vector spaces, without taking introductory courses about linear algebra to be rigorous from start. Any way I stumbled across
this proof

View attachment 31778

and my problem is in proof b.
View attachment 31779

cant he use vs 6. what prevents him from doing so.
then (-a)x = (-1 * a)x = -(ax) = (a * -1)x = a(-x)
what is error in my proof so he didnt use it
Click to expand...
But he did use it. That's the last line of his proof of (b):

​
1647985771751.png​

But in order to do so, you have to prove that (-1)x = -x, which you implicitly use twice, and which is what the rest of his proof shows:

1647985811461.png​

So what's wrong in your proof is that you haven't shown that to be true, but have assumed it. (I'd have expected that this would be stated separately as a theorem, but apparently it is not.)
 
my question isnt c*v = cv if c is element of F. so by definition isnt (-1)v = -v. why do we need yo prove it.
 
AbdelRahmanShady said:
my question isn't c*v = cv if c is element of F. so by definition isn't (-1)v = -v. why do we need to prove it.
Click to expand...
That's what he's proving! So clearly it hasn't yet been proved in this book.

Please show me the definition that states this fact!

You say you wanted to be rigorous; you haven't learned what that means yet. Nothing can be assumed that has not been explicitly proved.
 
we are proving (-a)x = -(ax), why vs 6 is not enough to prove that. he doesnt want to prove additive inverse he just wants to add (-a)x = -(ax)
 
AbdelRahmanShady said:
View attachment 31784
this
Click to expand...
That definition (as far as you've shown it) doesn't even mention -v! How can you claim that the definition states that (-1)v = -v?

When you say "by definition" in math, it must be taken literally: the definition itself must state the fact. The phrase may be used a little more lightly in the real world, but not here.
AbdelRahmanShady said:
we are proving (-a)x = -(ax), why vs 6 is not enough to prove that. he doesnt want to prove additive inverse he just wants to add (-a)x = -(ax)
Click to expand...
VS6 is about scalar multiplication, not additive inverses.

In (-a)x = -(ax), -a is the additive inverse of a number (a), and -(ax) is the additive inverse of a vector (ax). Neither is a mere multiplication (in itself). You need to prove the relationship.

I think you just don't fully understand the nature of proof. Every statement must be based on a fact from a definition, axiom, or theorem. Relationships that seem obvious from our experiences with numbers can't be assumed to be true of a newly defined entity like vectors.
 
I am confused doesnt -v means -1 v or -v is a seperate entity that need to be proven to be equal to -1 v if so how do we know that -v + v = 0 if i cant say -v + v = (-1)v +1v = - (1-1)v = 0v = 0
 
Oh i see now. I didnt see it before but he define -x as addittive inverse so we neededd to prove that -1 x is the additive inverse. I firstly thought we wrote -x as short hand for -1 x as we used to do in regular algebra
 
AbdelRahmanShady said:
Oh i see now. I didnt see it before but he define -x as additive inverse so we neededd to prove that -1 x is the additive inverse. I firstly thought we wrote -x as short hand for -1 x as we used to do in regular algebra
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Ah! You were indeed assuming something from your past experience, as I figured, apparently because you hadn't even noticed the definition.

Just before you wrote this, I found a copy of your book, and saw that the definition I asked for, of -v, is given immediately before the theorem you are asking about:

1647992907245.png​

Until this point, "-x" had no meaning; now it is defined. Until the theorem, "-x" meant only the vector you add to x to obtain the 0 vector; after the theorem, we know it is also equal to (-1)x.

Nothing can be assumed as you did; there is no "shorthand" unless we define its meaning precisely. That is the meaning of "rigorous" in mathematics. Now you know.
 
topsquark said:
The point is, I think, that the additive inverse of the vector v, which is denoted as -v, has to at some point be shown to be equal to (-1)v. It's "obvious" but it must be proved.

-Dan
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I learned that nothing is obvious in math and that everything must be proved.
As a student I thought that my professor lost his mind by proving the intermediate value theorem in class! Of course, if you drive your car from 50 mph to 60mph you reached 52.5mph at some point. Everyone know that! I struggled with that one for a while as a student. I was wondering what I was doing being a math major where you have to prove the obvious. But I finally got it.

topsquark, I am not saying that I disagreed with anything that you said. In fact, you did say that the obvious must be proved.
 
One last thing, does this mean that for this proof to work -1 must be element of F which we implicitely assumed. Secondly does normal axioms works here. Like if x = y then can i say x + v = y + v without proof or do i to say x + v = x + v and since x = y rhs = y + v so x + v = y + v.
 
AbdelRahmanShady said:
does this mean that for this proof to work -1 must be element of F which we implicitely assumed.
Click to expand...
By definition, scalar multiplication involves a scalar, which is an element of the field F used in defining the vector space. How could you multiply by anything else? This is an explicit "assumption"!

AbdelRahmanShady said:
does normal axioms works here. Like if x = y then can i say x + v = y + v without proof or do i to say x + v = x + v and since x = y rhs = y + v so x + v = y + v.
Click to expand...
Since you are defining a new entity called a vector space, the only facts you can use are those that have been explicitly included in the definition (which, as presented to you, includes a set of axioms), or proved as theorems -- or are known already, such as the field axioms applied to F, and the properties of equality (an equivalence relation).

What you are asking about, essentially substitution, is implied by the definition of (vector) addition, namely that x+y is required to be uniquely defined. As you say, since any vector is equal to itself (a property of equality), and since adding the same thing to the same thing must result in the same thing (uniqueness), substitution must be valid.

I don't know whether this is explicitly stated in your book; it's possible that it is not, but your reasoning shows that it is valid.
 
What do you mean by substitution. Does it means replacing x with y. If so shouldnt it be always valid as x =y. Or there is case where it is not
 
AbdelRahmanShady said:
What do you mean by substitution. Does it means replacing x with y. If so shouldnt it be always valid as x =y. Or there is case where it is not
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I'm not sure why I used that word. I meant the property you are asking about, whatever you call it. (Some books call it the addition property of equality.)

Substitution is what you did in saying "since x = y rhs = y + v", which perhaps is what my mind was on.
 
Ok i thought substitution meant re[lacing x with y in any equation which should always be vslid as they are equal. Any way thanks tremendously for help as you cut hours of me cutting my head
 
One more thing, in my proof I said x+v = x+v thats because of uniqueness and that any vector is equal itself. Then when I replaced rhs with y + v thats because x = y and result of addition is unique so y + v is equal to x + v.
So in this proof we use equality once and uniqueness twice ,right?
 
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