Proof of linear independence

**Mollier** · January 14th 2010, 08:00 PM

problem:

(a)
Prove that the four vectors
$x=(1,0,0),\;y=(0,1,0),\;z=(0,0,1)\;u=(1,1,1)$ in $\mathbb{C}^3$ form a linearly dependent set, but any three of them are linearly independent.

(b)
$x(t)=1,\;y(t)=t,\;z(t)=t^2,\;u(t)=1+t+t^2$ , prove that $x,y,z,u$ are linearly dependent, but any three of them are linearly independent.

attempt:

(a) To prove that the four vectors are linearly dependent, I guess I could set up four linear equations and solve them.
But I'm thinking it should be enough to show that:
$x+y+z-u=0$ .

The thing I do not understand is the "any three of them are linearly independent" part.
How do I go about proving that without checking all combinations?

(b) $x+y+z-u=0$ . Again, I don't know how to prove the "any three of them" part without checking all combinations.

Thanks!

**Black** · January 14th 2010, 09:04 PM

For both (a) and (b), you're pretty much spot on for proving linear dependence.

For (a), x, y, and z being independent is clear. Therefore, we need to show that u and any two of x,y, and z are linearly independent. Let $x=\mathbf{e}_1, y=\mathbf{e}_2, z=\mathbf{e}_3$ and let $i,j \in \{1,2,3\}$ , where $i \not= j$ . If we have (for $a,b,c \in \mathbb{C}$ )

$a\mathbf{e}_i+b\mathbf{e}_j+cu=0$ ,

then no matter what i and j are, we will always end up with the following system of equations:

$a+c=0$

$b+c=0$

$c=0$ ,

which implies $a=b=c=0$ .

Working out part (b) is very similar.

**Mollier** · January 14th 2010, 10:37 PM

Originally Posted by Black

For (a), x, y, and z being independent is clear. Therefore, we need to show that u and any two of x,y, and z are linearly independent.

Can I use the statement that "x,y,z being independent is clear" in a proof?

As for (b), is this ok:

Let $x=t^0,\;y=t^1,\;z=t^2,\;i,j=\{0,1,2\},\;i\neq j$
If for $a,b,c \in \mathbb{C}$ :

$at^i+bt^j+cu=0$

then for all i and j we get a=b=c=0.

Thanks.

**Black** · January 15th 2010, 02:59 AM

Originally Posted by Mollier

Can I use the statement that "x,y,z being independent is clear" in a proof?

I guess it depends on the teacher that's grading your work. You should prove that they are independent (to be on the safe side), but it's very straightforward.

Originally Posted by Mollier

As for (b), is this ok:

Let $x=t^0,\;y=t^1,\;z=t^2,\;i,j=\{0,1,2\},\;i\neq j$
If for $a,b,c \in \mathbb{C}$ :

$at^i+bt^j+cu=0$

then for all i and j we get a=b=c=0.

Thanks.

Yep, pretty much. No matter what i and j are, when you group the like terms together and compare coefficients, you'll end up with the same system of equations as part (a).

**Raoh** · January 15th 2010, 03:40 AM

Originally Posted by Mollier

problem:

(a)
Prove that the four vectors
$x=(1,0,0),\;y=(0,1,0),\;z=(0,0,1)\;u=(1,1,1)$ in $\mathbb{C}^3$ form a linearly dependent set, but any three of them are linearly independent.

(b)
$x(t)=1,\;y(t)=t,\;z(t)=t^2,\;u(t)=1+t+t^2$ , prove that $x,y,z,u$ are linearly dependent, but any three of them are linearly independent.

attempt:

(a) To prove that the four vectors are linearly dependent, I guess I could set up four linear equations and solve them.
But I'm thinking it should be enough to show that:
$x+y+z-u=0$ .

The thing I do not understand is the "any three of them are linearly independent" part.
How do I go about proving that without checking all combinations?

(b) $x+y+z-u=0$ . Again, I don't know how to prove the "any three of them" part without checking all combinations.

Thanks!

hi
$u\in Span(z,y,x)$
and $(z,y,x)$ is a simple basis of $\mathbb{C}^3$ which means it's L.I.

**Raoh** · January 15th 2010, 03:47 AM

Originally Posted by Mollier

Can I use the statement that "x,y,z being independent is clear" in a proof?

As for (b), is this ok:

Let $x=t^0,\;y=t^1,\;z=t^2,\;i,j=\{0,1,2\},\;i\neq j$
If for $a,b,c \in \mathbb{C}$ :

$at^i+bt^j+cu=0$

then for all i and j we get a=b=c=0.

Thanks.

hi
put,
$\lambda _0+\lambda _1t+\lambda _2t^2=0$
For $t=0$ ,you get $\lambda _0$ .
Differentiate with respect to $t$ ,
$\lambda _1+2\lambda _2t=0.$
put $t=0$ and get $\lambda _1=0.$
....
...and you get $\lambda _2=0.$

**Mollier** · January 15th 2010, 03:57 AM

Great stuff guys, thank you very much!

Edit-
Roah: just saw you last post, sweet!

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