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Chapter 12: Problem 10

Prove Theorem 12.4: Let \(f\) be a symmetric bilinear form on \(V\) over \(K\) (where \(1+1 \neq 0\) ). Then \(V\) has a basis in which \(f\) is represented by a diagonal matrix.

Short Answer

Expert verified
To prove Theorem 12.4, we follow these steps: 1. Choose an arbitrary nonzero vector \(v_1\) in \(V\) and find a basis of \(V\) containing \(v_1\). Let this basis be \(B = \{v_1, w_2, w_3, \dots, w_n\}\). 2. Apply the Gram-Schmidt orthogonalization process on the basis \(B\) to obtain a new basis \(C=\{v_1, v_2, \dots, v_n\}\) for \(V\). 3. Show that \(C\) is an orthogonal basis for \(V\) with respect to the symmetric bilinear form \(f\), i.e., \(f(v_i, v_j) = 0\) for all \(i \neq j\). 4. Prove that in the basis \(C\), the symmetric bilinear form \(f\) is represented by a diagonal matrix, where the off-diagonal entries are zero, and the diagonal entries are given by \(f(v_i, v_i)\) for \(1 \leq i \leq n\). Thus, we conclude that \(V\) has a basis in which \(f\) is represented by a diagonal matrix.

Step by step solution

01

Define key terms and setup

Given a vector space \( V \) over a field \(K\) (where \(1+1 \neq 0\)), let \(f\) be a symmetric bilinear form on \(V\). We are tasked with showing that \(V\) has a basis in which \(f\) is represented by a diagonal matrix.
02

Identify an orthogonal basis for V

For a basis \(\{v_1, v_2, \dots, v_n\}\), if \(f(v_i, v_j) = 0\) for all \(i \neq j\), then it's called an orthogonal basis with respect to the symmetric bilinear form \(f\). Our goal will be to find an orthogonal basis for \(V\).
03

Choose an arbitrary nonzero vector

Select an arbitrary nonzero vector \(v_1\) in \(V\). Since \(f\) is a symmetric bilinear form, we can find a basis of \(V\) that contains \(v_1\). Let this basis of \(V\) be \(B = \{v_1, w_2, w_3, \dots, w_n\}\).
04

Apply Gram-Schmidt orthogonalization process

Now apply the Gram-Schmidt orthogonalization process on the basis \(B\). For each \(i\), we define a new vector \(v_i\) as follows: \(v_i = w_i - \sum_{j=1}^{i-1} a_{ij}v_j\) where \(a_{ij} = \frac{f(w_i, v_j)}{f(v_j, v_j)}\). By performing this process, we get a new basis \(C = \{v_1, v_2, \dots, v_n\}\) for \(V\).
05

Show that the new basis is orthogonal

Now let's show that the basis \(C\) is orthogonal with respect to \(f\). If you compute \(f(v_i, v_j)\) for any two distinct vectors \(v_i, v_j \in C\), then you should get: \( f(v_i,v_j) = f\left(w_i - \sum_{k=1}^{i-1} a_{ik}v_k,v_j\right) \) Using the properties of symmetric bilinear forms and linearity, we can write: \( f(v_i,v_j) = f(w_i,v_j) - \sum_{k=1}^{i-1} a_{ik}f(v_k,v_j) \) Since \(j < i\), we can determine \(a_{ij}\), and by substituting it, we will get: \( f(v_i,v_j) = f(w_i,v_j) - f(v_j,v_j)a_{ij} = 0 \) Therefore, \(C\) is an orthogonal basis for \(V\) with respect to the symmetric bilinear form \(f\).
06

Prove that f is represented by a diagonal matrix

Finally, let's show that in basis \(C = \{v_1, v_2, \dots, v_n\}\), the symmetric bilinear form \(f\) is represented by a diagonal matrix. Since \(C\) is an orthogonal basis, we have that \(f(v_i, v_j) = 0\) for all \(i \neq j\). Consequently, in the matrix representation of \(f\) with respect to the basis \(C\), all the off-diagonal entries will be zero. The diagonal entries of the matrix will be represented as \(f(v_i, v_i)\), for \(1 \leq i \leq n\). Thus, we have shown that \(V\) has a basis in which \(f\) is represented by a diagonal matrix, which completes the proof of Theorem 12.4.

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Most popular questions from this chapter

Let \(V\) be a vector space over \(K .\) A mapping \(f: \widehat{V \times V \times \ldots \times V} \rightarrow K\) is called a multilinear (or m-linear) form on \(V\) if \(f\) is linear in each variable; that is, for \(i=1, \ldots, m\) \\[f(\ldots, \overline{a u+b v}, \ldots)=a f(\ldots, \hat{u}, \ldots)+b f(\ldots, \hat{v}, \ldots)\\] where \(\ldots\) denotes the \(i\) th element, and other elements are held fixed. An \(m\) -linear form \(f\) is said to be alternating if \(f\left(v_{1}, \ldots v_{m}\right)=0\) whenever \(v_{i}=v_{j}\) for \(i \neq j .\) Prove the following: (a) The set \(B_{m}(V)\) of \(m\) -linear forms on \(V\) is a subspace of the vector space of functions from \(V \times V \times \cdots \times V\) into \(K\) (b) The set \(A_{m}(V)\) of alternating \(m\) -linear forms on \(V\) is a subspace of \(B_{m}(V)\)

For each of the following quadratic forms \(q(x, y, z),\) find a nonsingular linear substitution expressing the variables \(x, y, z\) in terms of variables \(r, s, t\) such that \(q(r, s, t)\) is diagonal: (a) \(q(x, y, z)=x^{2}+6 x y+8 y^{2}-4 x z+2 y z-9 z^{2}\) (b) \(q(x, y, z)=2 x^{2}-3 y^{2}+8 x z+12 y z+25 z^{2}\) (c) \(q(x, y, z)=x^{2}+2 x y+3 y^{2}+4 x z+8 y z+6 z^{2}\) In each case, find the rank and signature.

We say that \(B\) is Hermitian congruent to \(A\) if there exists a nonsingular matrix \(P\) such that \(B=P^{T} A \bar{P}\) or equivalently, if there exists a nonsingular matrix \(Q\) such that \(B=Q^{*} A Q .\) Show that Hermitian congruence is an equivalence relation. (Note: If \(P=\bar{Q}\), then \(P^{T} A \bar{P}=Q^{*} A Q\).)

Find the symmetric matrix that corresponds to each of the following quadratic forms: (a) \(q(x, y, z)=3 x^{2}+4 x y-y^{2}+8 x z-6 y z+z^{2}\) (b) \(q^{\prime}(x, y, z)=3 x^{2}+x z-2 y z\) (c) \(q^{\prime \prime}(x, y, z)=2 x^{2}-5 y^{2}-7 z^{2}\)

Let \(H=\left[\begin{array}{ccc}1 & 1+i & 2 i \\ 1-i & 4 & 2-3 i \\ -2 i & 2+3 i & 7\end{array}\right],\) a Hermitian matrix Find a nonsingular matrix \(P\) such that \(D=P^{T} H \bar{P}\) is diagonal. Also, find the signature of \(H\)
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