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Chapter 12: Problem 37

Give an example of a quadratic form \(q(x, y)\) such that \(q(u)=0\) and \(q(v)=0\) but \(q(u+v) \neq 0\)

Short Answer

Expert verified
The quadratic form \(q(x,y) = x^2 - y^2\) along with vectors \(u = (1, 1)\) and \(v = (1, -1)\) satisfy the given conditions, since \(q(u) = 0\), \(q(v) = 0\), and \(q(u+v) = 4 \neq 0\).

Step by step solution

01

Define a quadratic form

A quadratic form is essentially a homogeneous polynomial of degree 2. Let's define our quadratic form in two variables \( x \) and \( y \) as: \[ q(x, y) = x^2 - y^2 \]
02

Define the vectors \( u \) and \( v \)

Next, we need to choose two vectors \( u \) and \( v \) such that when we substitute them into \( q \), the result is zeros. Let's take \( u = (1, 1) \) and \( v = (1, -1) \). Let's substitute those vectors into the quadratic form \( q \): For \( u = (1, 1) \): The quadratic form gives \( q(u) = q(1,1) = 1^2 - 1^2 = 0 \). For \( v = (1, -1) \): Again, the quadratic form gives \( q(v) = q(1,-1) = 1^2 - (-1^2) = 0 \).
03

Calculate \( q(u + v) \)

Next, we need to check if \( q(u+v) \) is not equal to zero. Calculate \( u + v \): \( u + v = (1,1) + (1,-1) = (2,0) \). Substitute \( u + v \) into \( q \), i.e. \( q(u+v) = q(2,0) = 2^2 - 0^2 = 4 \). So, \( q(u + v) \neq 0 \). Hence, the quadratic form \( q \)= \( x^2 - y^2 \) along with vectors \( u = (1,1) \) and \( v = (1, -1) \) fulfill the conditions provided in the problem.

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Most popular questions from this chapter

Find the quadratic form \(q(X)\) that corresponds to each of the following symmetric matrices: (a) \(A=\left[\begin{array}{rr}5 & -3 \\ -3 & 8\end{array}\right]\) (b) \(B=\left[\begin{array}{rrr}4 & -5 & 7 \\ -5 & -6 & 8 \\ 7 & 8 & -9\end{array}\right]\) (c) \(C=\left[\begin{array}{rrrr}2 & 4 & -1 & 5 \\ 4 & -7 & -6 & 8 \\ -1 & -6 & 3 & 9 \\ 5 & 8 & 9 & 1\end{array}\right]\)

Let \(S(V)\) denote all symmetric bilinear forms on \(V\). Show that (b) If \(\operatorname{dim} V=n,\) then \(\operatorname{dim} S(V)=\frac{1}{2} n(n+1)\) (a) \(S(V)\) is a subspace of \(B(V)\)

Prove Theorem 12.7: Let \(f\) be a Hermitian form on \(V\). Then there is a basis \(S\) of \(V\) in which \(f\) is represented by a diagonal matrix, and every such diagonal representation has the same number \(\mathbf{p}\) of positive entries and the same number n of negative entries.

Consider the quadratic form \(q(x, y)=3 x^{2}+2 x y-y^{2}\) and the linear substitution \\[x=s-3 t, \quad y=2 s+t\\] (a) Rewrite \(q(x, y)\) in matrix notation, and find the matrix \(A\) representing \(q(x, y)\) (b) Rewrite the linear substitution using matrix notation, and find the matrix \(P\) corresponding to the substitution. (c) Find \(q(s, t)\) using direct substitution. (d) Find \(q(s, t)\) using matrix notation.

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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