YALinearAlgebraT: Proof of Linear Independence

[Ricemarine]

Problem in the book I'm confused with:

• {a, b, c, d} are linearly independent vectors.

• Would {a+b, b+c, c+d, d+a} be linearly independent or dependent?

So for a set of vectors to be linearly independent, the set of vectors must only have the trivial solution (Zero vector).

So I made a vector equation based on the times each variable was used and try and see if there is a nontrivial solution.

So
General solution...
a + b = 0
b + c = 0
c + d = 0
a + d = 0

which becomes
A B C D (hard to make every single matrix for each variable)
[ 1 1 0 0 | 0 ]
[ 0 1 1 0 | 0 ]
[ 0 0 1 1 | 0 ]
[ 1 0 0 1 | 0 ]

which when reduced becomes
[ 1 0 0 1 | 0 ]
[ 0 1 0 -1| 0 ]
[ 0 0 1 1 | 0 ]
[ 0 0 0 0 | 0 ]

Thus, since x4 is free (or d, not sure, someone tell me ), there exists weights not all zero that also makes them = 0...

But the book also states in as a warning that a general statement cannot be true just with a specific example... so I'm wondering if this works or not...


Thanks again for whoever helps .

 

[GoSharks]

Originally posted by: Ricemarine
But the book also states in as a warning that a general statement cannot be true just with a specific example... so I'm wondering if this works or not...

However, a statement can be shown to be false or not applicable if you can show a contradiction.

Statement: All dogs are red.
Example in support: Your dog is red.
contradiction: My dog is white.

Therefore, the statement is incorrect, and the contradiction is true: "All dogs are NOT red."

(Terms might be off by a bit - its been a year since discrete math, but the general idea still holds.)

 

[Born2bwire]

EDIT: Wait, I'm just rambling.

It's obvious that they're dependent because you can derive one of the vectors as a combination of the other three.

 

[chuckywang]

Answer: They're dependent.

Here is an example.
Let a = (1,0,0,0)
Let b = (0,1,0,0)
Let c = (0,0,1,0)
Let d = (0,0,0,1)

Obviously, a, b, c, and d are linearly independent.

However (1,1,0,0), (0,1,1,0), (0,0,1,1), and (1,0,0,1) are not linearly independent.

 

[silverpig]

Would {a+b, b+c, c+d, d+a} be linearly independent or dependent?

They're dependent.

let
w = a+b
x = b+c
y = c+d
z = d+a

You can write

w - x + y = a + b - b - c + c + d = a + d = z

So w-x+y=z, one of the vectors can be written as a sum of the others, and thus the set is not linearly independent.

 

[Ricemarine]

Originally posted by: GoSharks

Originally posted by: Ricemarine
But the book also states in as a warning that a general statement cannot be true just with a specific example... so I'm wondering if this works or not...

However, a statement can be shown to be false or not applicable if you can show a contradiction.

Statement: All dogs are red.
Example in support: Your dog is red.
contradiction: My dog is white.

Therefore, the statement is incorrect, and the contradiction is true: "All dogs are NOT red."

(Terms might be off by a bit - its been a year since discrete math, but the general idea still holds.)

Ok, so how do I justify this... that's the hardest part of this...

let
w = a+b
x = b+c
y = c+d
z = d+a

You can write

w - x + y = a + b - b - c + c + d = a + d = z

So w-x+y=z, one of the vectors can be written as a sum of the others, and thus the set is not linearly independent.

Isn't that a theorem (trying to remember)??? Is there also a legit way to find that linear combination without my teacher thinking "oh... he guess and checked... +0."???

Edit: Actually, my matrix if I didn't solve for the zero vector would give me weights that would tell me what you are posting above!!!.... But am I allowed to do that with a coefficient matrix (thinking... some doubt, anyone know?)... I know you can reduce, but usually they're with augmented matrices...