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Chapter 6: Problem 9

Let U be a linear operator on a finite-dimensional inner product space V. If \(\|\mathrm{U}(x)\|=\|x\|\) for all \(x\) in some orthonormal basis for \(\mathrm{V}\), must U be unitary? Justify your answer with a proof or a counterexample.

Short Answer

Expert verified
Yes, U must be unitary. Given that \(\|U(x)\| = \|x\|\) for all x in some orthonormal basis for V, we showed that the inner product is preserved for all x and y in V, i.e., \ = \ for all x and y in V. This confirms that U is unitary.

Step by step solution

01

Definition and properties of unitary operators

A linear operator U on an inner product space V is called unitary if it preserves the inner product, i.e., \ = \ for all x and y in V. Equivalently, U is unitary if its adjoint operator \(U^*\) satisfies the condition \(U^*U = UU^* = I\), where I is the identity operator. An important property of unitary operators is that they preserve the norm of vectors, i.e., \(\|U(x)\| = \|x\|\) for all x in V. Now let's use this information to check whether U must be unitary.
02

Work with given property

We are given that \(\|U(x)\| = \|x\|\) for all x in some orthonormal basis for V. Let's denote this basis by \(\{e_1, e_2, ..., e_n\}\), where n is the dimension of V. Now we want to check whether the inner product is preserved for this U. Let x and y be any vectors in V. Then, they can be represented using the orthonormal basis as follows: \[x = \sum_{i=1}^n a_ie_i\] \[y = \sum_{j=1}^n b_je_j\] where \(a_i\) and \(b_j\) are scalar coefficients.
03

Compute the inner products

Next, we will compute the inner products \ and \: \ = \ = \<(\(\sum_{i=1}^n a_iU(e_i)\)),\((\sum_{j=1}^n b_jU(e_j\))\) Using the bilinearity of the inner product, we get: \ = \(\sum_{i,j}a_ib_j\\) Now let's compute the inner product \: \ = \<(\(\sum_{i=1}^n a_ie_i),(\(\sum_{j=1}^n b_je_j\)\) By the bilinearity of the inner product, we get: \ = \(\sum_{i,j}a_ib_j\\)
04

Compare the inner products

Now, let us compare the expressions for \ and \. Notice that both expressions involve the same summation, \(\sum_{i,j}a_ib_j\). To show that U is unitary, we need to show that \ = \ for all x and y. To do this, it suffices to show that \ = \ for all i and j, because then the two expressions will be identical. Recall that we are given that \(\|U(x)\| = \|x\|\) for all x in the orthonormal basis for V. In particular, this implies \(\|U(e_i)\| = \|e_i\|\) for all i. Thus, we have: \(\|U(e_i)\|^2 = \|e_i\|^2\) \ = \ Using the properties of the inner product, we know that the left-hand side of the equation is the sum of the squares of the inner products of all pairs of vectors U(e_i) and U(e_j) with i = j. Similarly, the right-hand side is the sum of the squares of the inner products of all pairs of vectors e_i and e_j with i = j. Since the norms are preserved for the orthonormal basis vectors, it follows that \ = \ for all i and j.
05

Conclusion

Since \ = \ for all x and y in V, we can conclude that U is unitary. Consequently, if \(\|U(x)\| = \|x\|\) for all x in some orthonormal basis for V, then U must be unitary.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Operator
A linear operator is an important concept in linear algebra. It is essentially a function that maps elements from one vector space to another while preserving the operations of vector addition and scalar multiplication.
Imagine you have a structure full of vectors or objects that can be transformed, moved, or changed in some way. A linear operator allows you to perform these transformations in a consistent manner.
  • Additivity: For two vectors, say \( u \) and \( v \), a linear operator \( T \) must satisfy \( T(u + v) = T(u) + T(v) \).
  • Homogeneity: For any scalar \( c \) and vector \( u \), the linear operator will maintain \( T(cu) = cT(u) \).
These properties ensure that when a linear operator acts on a vector space, it does so predictably and consistently. Thus, you can think of it as a systematic "rule" for transforming vectors.
Inner Product Space
An inner product space is a vector space with an additional structure called an inner product. The inner product is a way to multiply vectors together, resulting in a scalar. This concept is crucial as it introduces geometric and algebraic properties into vector spaces.
Think of the inner product as a measure of how "aligned" two vectors are. For example:
  • Common notations for the inner product include \( \langle x, y \rangle \).
  • It satisfies certain conditions like commutativity: \( \langle x, y \rangle = \langle y, x \rangle \).
  • Linearity in the first argument: \( \langle ax + by, z \rangle = a\langle x, z \rangle + b\langle y, z \rangle \)
The concept of distance and angles between vectors comes from this inner product structure. For instance, the angle between two vectors and their length or magnitude is derived from this, making it a fundamental concept in understanding vector spaces and their transformations.
Orthonormal Basis
An orthonormal basis for an inner product space is a set of vectors that are both orthogonal and normalized. This means each vector in the set is orthogonal to every other vector, and each has a length of one. Such bases are particularly useful in simplifying problems in linear algebra.
Imagine having a perfect grid or a coordinate system where each direction is well-defined and independent of others. An orthonormal basis provides this clean structure.
  • Orthogonality: Any two different vectors, \( e_i \) and \( e_j \), in the basis satisfy \( \langle e_i, e_j \rangle = 0 \).
  • Normalization: Each vector in the basis has a unit length, \( \langle e_i, e_i \rangle = 1 \).
  • Representation: Any vector in the space can be expressed uniquely as a combination of the basis vectors.
Using an orthonormal basis can be extremely efficient for numerical computations, allowing for easy computation of projections and simplifications of equations involving linear operators and inner product spaces.

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Most popular questions from this chapter

Assume that \(\mathrm{T}\) is a linear operator on a complex (not necessarily finitedimensional) inner product space \(\mathrm{V}\) with an adjoint \(\mathrm{T}^{*}\). Prove the following results. (a) If \(\mathrm{T}\) is self-adjoint, then \(\langle\mathrm{T}(x), x\rangle\) is real for all \(x \in \mathrm{V}\). (b) If \(\mathrm{T}\) satisfies \(\langle\mathrm{T}(x), x\rangle=0\) for all \(x \in \mathrm{V}\), then \(\mathrm{T}=\mathrm{T}_{0}\). Hint: Replace \(x\) by \(x+y\) and then by \(x+i y\), and expand the resulting inner products. (c) If \(\langle\mathrm{T}(x), x\rangle\) is real for all \(x \in \mathrm{V}\), then \(\mathrm{T}\) is self-adjoint.

Let \(\mathrm{T}\) and \(\mathrm{U}\) be positive definite operators on an inner product space V. Prove the following results. (a) \(\mathrm{T}+\mathrm{U}\) is positive definite. (b) If \(c>0\), then \(c \mathrm{~T}\) is positive definite. (c) \(\mathrm{T}^{-1}\) is positive definite. Visit goo.gl/cQch7i for a solution.

Prove the Cayley-Hamilton theorem for a complex \(n \times n\) matrix \(A\). That is, if \(f(t)\) is the characteristic polynomial of \(A\), prove that \(f(A)=O\). Hint: Use Schur's theorem to show that \(A\) may be assumed to be upper triangular, in which case $$ f(t)=\prod_{i=1}^{n}\left(A_{i i}-t\right) . $$ Now if \(\mathrm{T}=\mathrm{L}_{A}\), we have \(\left(A_{j j} \mid-\mathrm{T}\right)\left(e_{j}\right) \in \operatorname{span}\left(\left\\{e_{1}, e_{2}, \ldots, e_{j-1}\right\\}\right)\) for \(j \geq 2\), where \(\left\\{e_{1}, e_{2}, \ldots, e_{n}\right\\}\) is the standard ordered basis for \(\mathrm{C}^{n}\). (The general case is proved in Section 5.4.)

In each part, apply the Gram-Schmidt process to the given subset \(S\) of the inner product space \(\mathrm{V}\) to obtain an orthogonal basis for \(\operatorname{span}(S)\). Then normalize the vectors in this basis to obtain an orthonormal basis \(\beta\) for \(\operatorname{span}(S)\), and compute the Fourier coefficients of the given vector relative to \(\beta\). Finally, use Theorem \(6.5\) to verify your result. (a) \(\mathrm{V}=\mathrm{R}^{3}, S=\\{(1,0,1),(0,1,1),(1,3,3)\\}\), and \(x=(1,1,2)\) (b) \(\mathbf{V}=\mathbf{R}^{3}, S=\\{(1,1,1),(0,1,1),(0,0,1)\\}\), and \(x=(1,0,1)\) (c) \(\mathrm{V}=\mathrm{P}_{2}(R)\) with the inner product \(\langle f(x), g(x)\rangle=\int_{0}^{1} f(t) g(t) d t\), \(S=\left\\{1, x, x^{2}\right\\}\), and \(h(x)=1+x\) (d) \(\mathrm{V}=\operatorname{span}(S)\), where \(S=\\{(1, i, 0),(1-i, 2,4 i)\\}\), and \(x=(3+i, 4 i,-4)\) (e) \(\mathrm{V}=\mathrm{R}^{4}, S=\\{(2,-1,-2,4),(-2,1,-5,5),(-1,3,7,11)\\}\), and \(x=\) \((-11,8,-4,18)\) (f) \(\mathbf{V}=\mathbf{R}^{4}, S=\\{(1,-2,-1,3),(3,6,3,-1),(1,4,2,8)\\}\), and \(x=(-1,2,1,1)\) (g) \(\mathrm{V}=\mathrm{M}_{2 \times 2}(R), S=\left\\{\left(\begin{array}{rr}3 & 5 \\ -1 & 1\end{array}\right),\left(\begin{array}{rr}-1 & 9 \\ 5 & -1\end{array}\right),\left(\begin{array}{rr}7 & -17 \\ 2 & -6\end{array}\right)\right\\}\), and \(A=\left(\begin{array}{rr}-1 & 27 \\ -4 & 8\end{array}\right)\) (h) \(\mathrm{V}=\mathrm{M}_{2 \times 2}(R), S=\left\\{\left(\begin{array}{ll}2 & 2 \\ 2 & 1\end{array}\right),\left(\begin{array}{rr}11 & 4 \\ 2 & 5\end{array}\right),\left(\begin{array}{ll}4 & -12 \\ 3 & -16\end{array}\right)\right\\}\), and \(A=\) \(\left(\begin{array}{rr}8 & 6 \\ 25 & -13\end{array}\right)\) (i) \(\mathrm{V}=\operatorname{span}(S)\) with the inner product \(\langle f, g\rangle=\int_{0}^{\pi} f(t) g(t) d t\), \(S=\\{\sin t, \cos t, 1, t\\}\), and \(h(t)=2 t+1\) (j) \(\mathrm{V}=\mathrm{C}^{4}, S=\\{(1, i, 2-i,-1),(2+3 i, 3 i, 1-i, 2 i)\), \((-1+7 i, 6+10 i, 11-4 i, 3+4 i)\\}\), and \(x=(-2+7 i, 6+9 i, 9-3 i, 4+4 i)\) (k) \(\mathrm{V}=\mathrm{C}^{4}, S=\\{(-4,3-2 i, i, 1-4 i)\), $$ (-1-5 i, 5-4 i,-3+5 i, 7-2 i),(-27-i,-7-6 i,-15+25 i,-7-6 i)\\}, $$ and \(x=(-13-7 i,-12+3 i,-39-11 i,-26+5 i)\) (1) \(\mathrm{V}=\mathrm{M}_{2 \times 2}(C), S=\left\\{\begin{array}{cc}1-i & -2-3 i \\ 2+2 i & 4+i\end{array}\right),\left(\begin{array}{cc}8 i & 4 \\\ -3-3 i & -4+4 i\end{array}\right)\), $$ \left.\left(\begin{array}{cc} -25-38 i & -2-13 i \\ 12-78 i & -7+24 i \end{array}\right)\right\\}, \text { and } A=\left(\begin{array}{rr} -2+8 i & -13+i \\ 10-10 i & 9-9 i \end{array}\right) $$ (m) \(\mathrm{V}=\mathrm{M}_{2 \times 2}(C), S=\left\\{\left(\begin{array}{cc}-1+i & -i \\ 2-i & 1+3 i\end{array}\right),\left(\begin{array}{cc}-1-7 i & -9-8 i \\ 1+10 i & -6-2 i\end{array}\right),\right.\), \(\left.\left(\begin{array}{cc}-11-132 i & -34-31 i \\ 7-126 i & -71-5 i\end{array}\right)\right\\}\), and \(A=\left(\begin{array}{cc}-7+5 i & 3+18 i \\ 9-6 i & -3+7 i\end{array}\right)\)

Label the following statements as true or false. (a) An inner product is a scalar-valued function on the set of ordered pairs of vectors. (b) An inner product space must be over the field of real or complex numbers. (c) An inner product is linear in both components. (d ) There is exactly one inner product on the vector space Rn. (e) The triangle inequality only holds in finite-dimensional inner product spaces. (f) Only square matrices have a conjugate-transpose. (g) If x, y, and z are vectors in an inner product space such that (x, y) = (x, z), then y = z. (h ) If (x, y) = 0 for all x in an inner product space. then y = 0 .
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