/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Let \(q\) be the quadratic form ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 11

Let \(q\) be the quadratic form associated with the symmetric bilinear form \(f\). Verify the polar identity \(\left.f(u, v)=\frac{1}{2}[q(u+v)-q(u)-q(v)] . \text { (Assume that } 1+1 \neq 0 .\right)\)

Short Answer

Expert verified
The polar identity for the quadratic form \(q\) and symmetric bilinear form \(f\) is verified as follows: 1. Expand the expression on the right-hand side: \( \frac{1}{2}[q(u+v)-q(u)-q(v)] = \frac{1}{2}[f(u+v, u+v) - f(u, u) - f(v, v)] \) 2. Use the properties of symmetric bilinear forms: \( \frac{1}{2}[q(u+v)-q(u)-q(v)] = \frac{1}{2}[f(u, v) + f(v, u)] \) 3. Since \(f(u, v) = f(v, u)\), we have: \( f(u, v) = \frac{1}{2}[q(u+v)-q(u)-q(v)] \) Thus, the polar identity is verified.

Step by step solution

01

Recall properties of symmetric bilinear forms and quadratic forms

A bilinear form is a function that takes a pair of vectors (u, v) and returns a scalar value. The form is symmetric if \(f(u, v) = f(v, u)\) for all u, v. A quadratic form q is associated with a symmetric bilinear form f. For any vector u, the quadratic form q(u) is equal to \(f(u, u)\).
02

Expand the expression on the right-hand side of the polar identity

Expand the quadratic forms in the expression on the right-hand side: \( \frac{1}{2}[q(u+v)-q(u)-q(v)] = \frac{1}{2} [f(u+v, u+v) - f(u, u) - f(v, v)] \)
03

Use the properties of symmetric bilinear forms and quadratic forms to show the left-hand side and the right-hand side are equal

Consider the properties of symmetric bilinear forms: \(f(u+v, u+v) = f(u, u) + f(u, v) + f(v, u) + f(v, v)\) Then, the expression on the right-hand side becomes: \( \frac{1}{2} [f(u+v, u+v) - f(u, u) - f(v, v)] = \frac{1}{2} [f(u, u) + f(u, v) + f(v, u) + f(v, v) - f(u, u) - f(v, v)] \) Simplify the expression on the right-hand side: \( \frac{1}{2} [f(u+v, u+v) - f(u, u) - f(v, v)] = \frac{1}{2} [f(u, v) + f(v, u)] \) Using the property that the bilinear form f is symmetric, we have: \(f(u, v) = f(v, u)\) Substitute this into the expression: \( \frac{1}{2} [f(u, v) + f(v, u)] = f(u, v) \) Now, we have shown that: \( f(u, v) = \frac{1}{2}[q(u+v)-q(u)-q(v)] \) Thus, we have verified the polar identity for the quadratic form q and symmetric bilinear form f.

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Most popular questions from this chapter

Let \([f]\) denote the matrix representation of a bilinear form \(f\) on \(V\) relative to a basis \(\left\\{u_{i}\right\\} .\) Show that the mapping \(f \mapsto[f]\) is an isomorphism of \(B(V)\) onto the vector space \(V\) of \(n\) -square matrices.

Let \(f\) be the bilinear form on \(\mathbf{R}^{2}\) defined by $$f\left[\left(x_{1}, x_{2}\right),\left(y_{1}, y_{2}\right)\right]=3 x_{1} y_{1}-2 x_{1} y_{2}+4 x_{2} y_{1}-x_{2} y_{2}$$ (a) Find the matrix \(A\) of \(f\) in the basis \(\left\\{u_{1}=(1,1), u_{2}=(1,2)\right\\}\) (b) Find the matrix \(B\) of \(f\) in the basis \(\left\\{v_{1}=(1,-1), \quad v_{2}=(3,1)\right\\}\) (c) Find the change-of-basis matrix \(P\) from \(\left\\{u_{i}\right\\}\) to \(\left\\{v_{i}\right\\},\) and verify that \(B=P^{T} A P\)

Show that \(q(x, y)=a x^{2}+b x y+c y^{2}\) is positive definite if and only if \(a>0\) and the discriminant \(D=b^{2}-4 a c<0\)

Show that congruence of matrices (denoted by \(\simeq\) ) is an equivalence relation; that is, (i) \(A \simeq A\) (ii) If \(A \simeq B,\) then \(B \simeq A\) (iii) If \(A \simeq B\) and \(B \simeq C,\) then \(A \simeq C\)

Consider a real quadratic polynomial \(q\left(x_{1}, \ldots, x_{n}\right)=\sum_{i, j=1}^{n} a_{i j} x_{i} x_{j},\) where \(a_{i j}=a_{j i}\) (a) If \(a_{11} \neq 0,\) show that the substitution \(x_{1}=y_{1}-\frac{1}{a_{11}}\left(a_{12} y_{2}+\cdots+a_{1 n} y_{n}\right), \quad x_{2}=y_{2}, \quad \ldots, \quad x_{n}=y_{n}\) yields the equation \(q\left(x_{1}, \ldots, x_{n}\right)=a_{11} y_{1}^{2}+q^{\prime}\left(y_{2}, \ldots, y_{n}\right),\) where \(q^{\prime}\) is also a quadratic polynomial. (b) If \(a_{11}=0\) but, say, \(a_{12} \neq 0,\) show that the substitution \(x_{1}=y_{1}+y_{2}, \quad x_{2}=y_{1}-y_{2}, \quad x_{3}=y_{3}, \quad \ldots, \quad x_{n}=y_{n}\) yields the equation \(q\left(x_{1}, \ldots, x_{n}\right)=\sum b_{i j} y_{i} y_{j},\) where \(b_{11} \neq 0,\) which reduces this case to case (a).
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