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Chapter 7: Problem 62

Suppose \(|\langle u, v\rangle|=\|u\|\|v\|\). (That is, the Cauchy-Schwarz inequality reduces to an equality.) Show that \(u\) and \(v\) are linearly dependent.

Short Answer

Expert verified
Given that \(|\langle u, v \rangle| = \|u\|\|v\|\), we can rewrite the equation as \((u_1v_1 + u_2v_2 + \cdots + u_nv_n)^2 = (u_1^2 + u_2^2 + \cdots + u_n^2)(v_1^2 + v_2^2 + \cdots + v_n^2)\). Using Lagrange's Identity, this equation is true if there is a non-zero scalar \(k\) such that \(u_1 = kv_1, u_2 = kv_2, \dots, u_n = kv_n\) for all \(i\). This implies that the vectors \(u\) and \(v\) are linearly dependent.

Step by step solution

01

Write down the Cauchy-Schwarz Inequality with equality#for_units#

The Cauchy-Schwarz inequality states that \( |\langle u, v \rangle| \leq \|u\|\|v\| \) for any vectors \(u\) and \(v\). In this problem, we are given that the inequality turns into an equality: \( |\langle u, v \rangle| = \|u\|\|v\| \)
02

Use the definition of the dot product#for_units#

Recall that the dot product of \(u = (u_1, u_2,\dots, u_n)\) and \(v = (v_1, v_2,\dots, v_n)\) can be represented as: \( \langle u, v \rangle = u_1v_1 + u_2v_2 + \cdots + u_nv_n \) So, we have: \( |\langle u, v \rangle| = |u_1v_1 + u_2v_2 + \cdots + u_nv_n| \)
03

Square both sides#for_units#

In order to simplify the expression and eliminate the absolute value, let's square both sides of the equation: \( |\langle u, v \rangle|^2 = (\|u\|\|v\|)^2 \)
04

Substitute the expressions for dot product and magnitudes#for_units#

Recall the expressions for the dot product and the magnitudes of vectors \(u\) and \(v\): \( \langle u, v \rangle = u_1v_1 + u_2v_2 + \cdots + u_nv_n \) \( \|u\|^2 = u_1^2 + u_2^2 + \cdots + u_n^2 \) \( \|v\|^2 = v_1^2 + v_2^2 + \cdots + v_n^2 \) Now substitute these expressions into the equation obtained in Step 3: \((u_1v_1 + u_2v_2 + \cdots + u_nv_n)^2 = (u_1^2 + u_2^2 + \cdots + u_n^2)(v_1^2 + v_2^2 + \cdots + v_n^2) \)
05

Show that u and v are linearly dependent#for_units#

From Step 4, we have the equation: \((u_1v_1 + u_2v_2 + \cdots + u_nv_n)^2 = (u_1^2 + u_2^2 + \cdots + u_n^2)(v_1^2 + v_2^2 + \cdots + v_n^2) \) We can rewrite this equation as follows: \((u_1v_1 + u_2v_2 + \cdots + u_nv_n)^2 - (u_1^2 + u_2^2 + \cdots + u_n^2)(v_1^2 + v_2^2 + \cdots + v_n^2) = 0 \) This last equation is an example of a well-known identity (which can be proved using induction), called Lagrange's Identity: \( a_1^2b_1^2 + a_2^2b_2^2 + \cdots + a_n^2b_n^2 - (a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2 =0 \) Lagrange's Identity states that the equation is true if and only if there is a non-zero scalar \(k\) such that \(a_i = kb_i\) for all \(i\), that is: \(u_1 = kv_1, u_2 = kv_2, \dots, u_n = kv_n\) Since we have shown that there exists a non-zero scalar \(k\) such that the components of \(u\) are proportional to the components of \(v\), we can conclude that the vectors \(u\) and \(v\) are linearly dependent.

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Most popular questions from this chapter

Find the minimal polynomial of each of the following matrices. (a) \(\left(\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\) (c) $\left(\begin{array}{rrr}4 & -14 & 5 \\ 1 & -4 & 2 \\ 1 & -6 & 4\end{array}\right)$ (d) $\left(\begin{array}{rrr}3 & 0 & 1 \\ 2 & 2 & 2 \\ -1 & 0 & 1\end{array}\right)$

Let \(S\) consist of the following vectors in \(\mathbf{R}^{4}\) : \\[u_{1}=(1,1,1,1), \quad u_{2}=(1,1,-1,-1), \quad u_{3}=(1,-1,1,-1), \quad u_{4}=(1,-1,-1,1)\\] (a) Show that \(S\) is orthogonal and a basis of \(\mathbf{R}^{4}\). (b) Write \(v=(1,3,-5,6)\) as a linear combination of \(u_{1}, u_{2}, u_{3}, u_{4}\) (c) Find the coordinates of an arbitrary vector \(v=(a, b, c, d)\) in \(\mathbf{R}^{4}\) relative to the basis \(\mathcal{S}\) (d) Normalize \(S\) to obtain an orthonormal basis of \(\mathbf{R}^{4}\).

Prove Theorem 7.17: Let \(A\) be the matrix representation of any inner product on \(V\). Then \(A\) is a positive definite matrix. Because \(\left\langle w_{i}, w_{j}\right\rangle=\left\langle w_{j}, w_{i}\right\rangle\) for any basis vectors \(w_{i}\) and \(w_{j},\) the matrix \(A\) is symmetric. Let \(X\) be any nonzero vector in \(\mathbf{R}^{n}\). Then \([u]=X\) for some nonzero vector \(u \in V\). Theorem 7.16 tells us that \(X^{T} A X=[u]^{T} A[u]=\langle u, u\rangle>0 .\) Thus, \(A\) is positive definite.

Let \(A\) be an \(n \times n\) matrix whose characteristic polynomial splits. Prove that \(A\) and \(A^{t}\) have the same Jordan canonical form, and conclude that \(A\) and \(A^{t}\) are similar. Hint: For any eigenvalue \(\lambda\) of \(A\) and \(A^{t}\) and any positive integer \(r\), show that $\operatorname{rank}\left((A-\lambda I)^{r}\right)=\operatorname{rank}\left(\left(A^{t}-\lambda I\right)^{r}\right) .$

Let \(V\) be a vector space and \(\beta_{1}, \beta_{2}, \ldots, \beta_{k}\) be disjoint subsets of \(V\) whose union is a basis for \(\mathrm{V}\). Now suppose that \(\gamma_{1}, \gamma_{2}, \ldots, \gamma_{k}\) are linearly independent subsets of \(\mathrm{V}\) such that \(\operatorname{span}\left(\gamma_{i}\right)=\operatorname{span}\left(\beta_{i}\right)\) for all \(i\). Prove that $\gamma_{1} \cup \gamma_{2} \cup \cdots \cup \gamma_{k}\( is also a basis for \)\mathrm{V}$.
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