# Thread: URGENT: Subspaces spanned by independent vectors in an inner product space.

1. ## URGENT: Subspaces spanned by independent vectors in an inner product space.

I have no idea where to go with this.

Show that if x1,...,xn are independent vectors in an inner product space, and y≠ 0 is such that y is orthogonal to xi for each i, then y is not a linear combination of x1,...,xn (i.e., is not in the subspace spanned by x1,...,xn )

I know that the x's being independent means that a1x1 + a2x2+...+anxn = 0 iff a1=a2=...=an=0

and that <y,xi> = 0 for each i

And i know that we have to show by the contrapositive that y ≠ a linear combination of the x's so we assume

y = b1x1 + ... + bnxn

I just dont know where to go from here.

2. Originally Posted by Mandi_Moo
I have no idea where to go with this.

Show that if x1,...,xn are independent vectors in an inner product space, and y≠ 0 is such that y is orthogonal to xi for each i, then y is not a linear combination of x1,...,xn (i.e., is not in the subspace spanned by x1,...,xn )

I know that the x's being independent means that a1x1 + a2x2+...+anxn = 0 iff a1=a2=...=an=0

and that <y,xi> = 0 for each i

And i know that we have to show by the contrapositive that y ≠ a linear combination of the x's so we assume

y = b1x1 + ... + bnxn

I just dont know where to go from here.
Hint: If $y = b_1x_1 + \ldots + b_nx_n$ then $\langle y,y\rangle = \langle b_1x_1 + \ldots + b_nx_n,y\rangle$.

(The information about $x_1,\,\ldots\,,x_n$ being independent is irrelevant. The result is still true if these vectors are not independent.)

3. So after i do that i get <y,y> = 0 which i have said implys that y is orthogonal to y, which dosent make sense so our assumption that y = b1*x1..... is false and b is not a linear combination of x1,...,xn?

4. Originally Posted by Mandi_Moo
So after i do that i get <y,y> = 0 which i have said implys that y is orthogonal to y, which dosent make sense so our assumption that y = b1*x1..... is false and b is not a linear combination of x1,...,xn?
That's right: <y,y> = 0 implies that y = 0, which contradicts the initial information that y ≠ 0.