If A(x,y) is a Positive Definate Bilinear Form then sqrt(A(x,x)) defines a norm

[deluks917]

Homework Statement

The Problem is from Mendelson Topology. Let V be a vector field with the real numbers as scalars. He defines a bilinear form as a function A:V x V -> R s.t for all x,y,z an element of V and real numbers a,b,c A(ax +by, z) = aA(x,z) + bA(y,z) and A(x,by + cz) = bA(x,y) + cA(x,z). A bilinear form is positive definate if A(x,x) > 0 for all x inequal to the zero vector. The question is to show that N(x) = sqrt(A(x,x)) defines a norm on V. I cannot seem to prove that the triangle inequality holds, that is N(x+y) <= N(x) + N(y).

Homework Equations

A function N(x) is a norm on V iff:
1) N(x) >= 0
2) N(x) = 0 iff x = zero vector
3) N(ax) = |a|N(x)
4) N(x + y) <= N(x) + N(y)

A being bilinear can be restated (I think):
A(x+y,z) = A(x,z) + A(y,z)
A(x,y+z) =A(z,y) + A(x,z)
A(ax,z) = aA(x,z)
A(x,az) = aA(x,z)

The Attempt at a Solution

The first three properties are easy to prove. I can't seem to show the fourth one. We seem to need:

sqrt(A(x,x) + A(x,y) + A(y,x) + A(y,y) ) <= sqrt(A(x,x)) + sqrt(A(y,y))

squaring both sides:
A(x,x) + A(x,y) + A(y,x) + A(y,y) <= A(x,x) + A(y,y) +2sqrt(A(x,x))sqrt(A(y,y)) or

A(x,y) + A(y,x) <= 2sqrt(A(x,x))*sqrt(A(y,y))

This is where I'm stuck. In Rⁿ the usual bi-linear form is, I think, the dot product. Hence this last property is just the Cauchy-Schwarz inequality. But I don't see how to do the problem in an arbitary vector space. Thanks in advance for any help.

 

[mighty2000]

Thank you for posting your question. I am a scientist and I would be happy to help you with this problem.

To prove the triangle inequality for N(x), we need to show that for any vectors x and y in V, N(x + y) <= N(x) + N(y). Let's start by expanding N(x + y) using the definition of N(x):

N(x + y) = sqrt(A(x + y, x + y))

Next, we can use the bilinear form property to expand A(x + y, x + y) as follows:

A(x + y, x + y) = A(x, x) + A(x, y) + A(y, x) + A(y, y)

Substituting this into our expression for N(x + y), we get:

N(x + y) = sqrt(A(x, x) + A(x, y) + A(y, x) + A(y, y))

Next, we can use the property of A(x, x) being positive definite to simplify this expression:

N(x + y) = sqrt(A(x, x) + 2A(x, y) + A(y, y))

Now, using the property of A(x, y) being bilinear, we can rewrite this as:

N(x + y) = sqrt(A(x, x) + 2A(x, y) + A(y, x) + A(y, y))

Next, we can use the property of A(x, y) being symmetric to rewrite this expression as:

N(x + y) = sqrt(A(x, x) + 2A(x, y) + 2A(x, y) + A(y, y))

Now, using the property of A(x, x) being positive definite again, we can simplify this expression to:

N(x + y) = sqrt(A(x, x) + 4A(x, y) + A(y, y))

Finally, we can use the Cauchy-Schwarz inequality to show that this expression is less than or equal to sqrt(A(x, x)) + sqrt(A(y, y)), which is equal to N(x) + N(y). Therefore, we have shown that N(x + y) <= N(x) + N(y), which proves the triangle inequality for N(x).

I hope this helps. Let me know if you have any further questions.[Your