Inner product and hermitic product

[damabo]

Homework Statement

prove the triangle inequality ||v+w|| ≤ ||v||+||w||
if (ℂ, V, + , [ , ] ) is a Hermitic space where [ , ] is the Hermitic product.

Homework Equations

|[v,w]| ≤ ||v|| . ||w||

$\overline{a+b i}$= a-b i (where i is the imaginary number, a+bi a complex number)

The Attempt at a Solution

||v+w||² = [v+w,v+w]. Since [v+w,v+w] = $\overline{[v+w,v+w]}$, [v+w,v+w] must be a real number. this means that the same rules will apply as in the inner product on the innerproductspace (ℝ, V, +, [ , ] ) ( true ?). so [v+w,v+w] = [v,v] + [w,w] + 2[v,w] ≤ ||v||²+||w||²+2||v|| ||w|| = (||v|| + ||w|| ) ²
this means ||v+w|| ≤ ||v|| + ||w||

 

[mighty2000]

To prove the triangle inequality, we will use the Hermitic product and the property that |[v,w]| ≤ ||v|| . ||w||. We will start by expanding ||v+w||² using the Hermitic product as follows:

||v+w||² = [v+w,v+w] = [v,v] + [v,w] + [w,v] + [w,w]

Since the Hermitic product is a real number, we can take its complex conjugate without changing its value. Therefore, we can rewrite the above equation as:

||v+w||² = \overline{[v,v] + [v,w] + [w,v] + [w,w]}

Using the property that |[v,w]| ≤ ||v|| . ||w||, we can further simplify the equation as:

||v+w||² = \overline{[v,v] + [w,w]} + \overline{[v,w] + [w,v]}

Next, we can use the property that \overline{a+b i}= a-b i to rewrite the above equation as:

||v+w||² = [v,v] + [w,w] + [v,w] + [w,v]

Now, we can use the Hermitic product to rewrite [v,w] and [w,v] as:

||v+w||² = [v,v] + [w,w] + [v,w] + [w,v] = [v,v] + [w,w] + 2[v,w]

Finally, we can use the property that |[v,w]| ≤ ||v|| . ||w|| to rewrite the above equation as:

||v+w||² = [v,v] + [w,w] + 2[v,w] ≤ ||v||² + ||w||² + 2||v|| ||w|| = (||v|| + ||w||)²

Since ||v+w||² is a real number, we can take its square root without changing its value. Therefore, we can conclude that:

||v+w|| ≤ ||v|| + ||w||

Thus, we have proven the triangle inequality for a Hermitic space.