## find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

Linear spaces and subspaces, linear transformations, bases, etc.

### find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

Find a basis for the set of all vectors of the form (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c).

This made so much sense in the book.
lawrence

Posts: 15
Joined: Wed Apr 01, 2009 10:48 pm

### Re: find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

$(a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)=a(1,2,-1,3)+b(-2,5,-4,1)+c(5,-8,7,1)$
Matt

Posts: 10
Joined: Mon May 18, 2009 3:30 pm

### Re: find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

So the basis is just $(1,2,-1,3), (-2,5,-4,1), (5,-8,7,1)$? It's that easy?
lawrence

Posts: 15
Joined: Wed Apr 01, 2009 10:48 pm

### Re: find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

lawrence wrote:So the basis is just $(1,2,-1,3), (-2,5,-4,1), (5,-8,7,1)$? It's that easy?

Martingale

Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

Could you give me another hint, please?
lawrence

Posts: 15
Joined: Wed Apr 01, 2009 10:48 pm

### Re: find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

$(1,2,-1,3)+(-2)\cdot(-2,5,-4,1)=(5,-8,7,1)$

therefore the vectors are not linearly independent

Martingale

Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

So this is where I have to do the vector-sum equals zero thing? When I solve the matrix, I get...

$\left[\begin{array}{cccc}1&0&1&0\\0&1&-2&0\\0&0&0&0\\0&0&0&0\end{array}\right]$

So a=-c, b=2c. If c=-1, then a=1,b=-2 and (1,2,-1,3)+(4,-10,8,-2)+(-5,8,-7,-1) = (0,0,0,0). So the answer is just the first two vectors, (1,2,-1,3),(-2,5,-4,1)?
lawrence

Posts: 15
Joined: Wed Apr 01, 2009 10:48 pm

### Re: find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

lawrence wrote:So this is where I have to do the vector-sum equals zero thing? When I solve the matrix, I get...

$\left[\begin{array}{cccc}1&0&1&0\\0&1&-2&0\\0&0&0&0\\0&0&0&0\end{array}\right]$

So a=-c, b=2c. If c=-1, then a=1,b=-2 and (1,2,-1,3)+(4,-10,8,-2)+(-5,8,-7,-1) = (0,0,0,0). So the answer is just the first two vectors, (1,2,-1,3),(-2,5,-4,1)?

those two vectors will work

Martingale

Posts: 350
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: find basis for (a-2b+5c,2a+5b-8c,-a-4b+7c,3a+b+c)

I don't know why I'm having so much trouble with this. Thank you for your help!
lawrence

Posts: 15
Joined: Wed Apr 01, 2009 10:48 pm