Math Is Fun Forum

GemmaJ1988

Member

Registered: 2008-10-08

Posts: 39

linear algebra

suppose

If ad-bc=0 show that v and w are linearly dependent

does anyone know how i do this.
i know to show something is linearly independent i would do as follows:

then try and solve the simultaneous eqns.
but seen as  v and w are linearly dependent im a little lost please help

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: linear algebra

You have your definitions a little backwards.  If you're able to find a pair (x, y) that satisfies your equation above then you've shown that they are linearly dependent.  Since that's what you're asked to prove, just find that (x, y) pair and you're done.

---

Wrap it in bacon

GemmaJ1988

Member

Registered: 2008-10-08

Posts: 39

Re: linear algebra

i can see that but i dont understand where i get ad-bc=0 from

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: linear algebra

You have 2 simultaneous equations: ax + cy = 0 and bx + dy = 0.  Try solving those 2 equations for x and y.  Along the way you'll need to use the fact that ad - bc = 0.

---

Wrap it in bacon

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: linear algebra

You won't exactly be able to "solve" for x and y, but you will be able to say:

0*y = 0 (or 0*x = 0)

Meaning that y (or x) can be any value.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

GemmaJ1988

Member

Registered: 2008-10-08

Posts: 39

Re: linear algebra

does anyone know where the ad-bc=0 comes from this is the bit thats really confusing me! i know that the vectors are linearly dependent i just dont know how to show this using ad-bc=0.

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: linear algebra

did you try what dude say?

ax + cy = 0
bx + dy = 0 -> multiply by c/d
bcx/d + cy = 0 -> subtract from first
ax-bcx/d = 0 -> multiply by d
adx - bcx = 0 -> divide by x
ad - bc = 0

---

The Beginning Of All Things To End.
The End Of All Things To Come.

GemmaJ1988

Member

Registered: 2008-10-08

Posts: 39

Re: linear algebra

Thank you for all your help I understand this now! I was clearly going about it the wrong way!

mathsmypassion

Member

Registered: 2008-12-01

Posts: 33

Re: linear algebra

Actually you can simplify the solution. Instead of trying to find x and y, you can look for only one variable z.
The two vectors are linearly dependent if  there is a real z so (a b) +z(c d)=0 . That means a+zc=0 and b+zd=0 or a/c=b/d or ad-bc=0