I've just been introduced to the concept of a bilinear form, and I'm having trouble understanding how addition is defined with 2 bilinear forms, which a question in a textbook asks:
Show the sum of 2 bilinear forms is also a bilinear form.
Trivial Problem: Sum of 2 bilinear forms is also a bilinear form
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linear-algebra
2 Answers
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A bilinear form is just a type of function. Therefore, addition of bilinear forms is just regular function addition: pointwise addition.
If $f, g$ are bilinear forms then
$$(f+g)(v, u) = f(v, u) + g(v, u)$$
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1Might be clearer to write $(f+g)(x,y) = f(x,y) + g(x,y)$. – 2017-01-10
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1@Batman agreed, thanks for pointing that out; edited accordingly – 2017-01-10
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1probably worth adding that, in finite dimension with chosen basis, we are just adding matrices; not necessarily symmetric if the form is not. – 2017-01-10
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For every $f,g: E\times F\to \mathbb{K}$ bilinear forms, for all $\lambda,\mu\in \mathbb{K},$ for all $x,y\in E$ and for all $z\in F$ $$(f+g)\left(\lambda x+\mu y,z\right)=f\left(\lambda x+\mu y,z\right)+g\left(\lambda x+\mu y,z\right)$$$$=\lambda f(x,z)+\mu f(y,z)+\lambda g(x,z)+\mu g(y,z)$$$$=\lambda \left(f(x,z)+g(x,z)\right)+\mu\left(f(y,z)+g(y,z)\right)$$$$=\lambda \left(f+g\right)(x,z)+\mu \left(f+g\right)(y,z).$$ In the same way, for all $\lambda,\mu\in \mathbb{K},$ for all $x\in E$ and for all $y,z\in F$ : $$(f+g)\left(x,\lambda y+\mu z\right)=\ldots=\lambda \left(f+g\right)(x,y)+\mu \left(f+g\right)(x,z).$$ So, $f+g: E\times F\to \mathbb{K}$ is a bilinear form.