Inner Product and Hilbert Spaces Questions and Answers

Q: Let \(H\) be a Hilbert space and \(S^{\perp}\) denote the orthogonal complement of a set \(S \subseteq H\). Which of the following is INCORRECT?

The incorrect statement is \(S_{1} \subseteq S_{2} \Rightarrow S_{1}^{\perp} \subseteq S_{2}^{\perp}\).Taking orthogonal complements reverses inclusion. A vector orthogonal to every element of the larger set \(S_{2}\) is orthogonal to every element of \(S_{1}\), so the true statement is \(S_{2}^{\perp} \subseteq S_{1}^{\perp}\). For example, \(\{0\} \subseteq H\) but \(\{0\}^{\perp} = H\) is not contained in \(H^{\perp} = \{0\}\).The other three are correct. Every \(s \in S\) is orthogonal to every element of \(S^{\perp}\), so \(S \subseteq (S^{\perp})^{\perp}\). Every vector is orthogonal to 0, so \(\{0\}^{\perp} = H\). And \(S^{\perp}\) is an intersection of kernels of the continuous functionals \(x \mapsto \langle x, s \rangle\), hence always closed.

Q: Let \(H = \{(x_{n}) \in \ell_{2} :\) \(\sum_{n=1}^{\infty} \frac{x_{n}}{n} = 1\}\). Then \(H\)

\(H\) is closed.Define \(\phi(x) = \sum \frac{x_{n}}{n}\) on \(\ell_{2}\). Since \((\frac{1}{n}) \in \ell_{2}\), the Cauchy-Schwarz inequality gives \(|\phi(x)| \le \|x\|_{2} \left(\sum \frac{1}{n^{2}}\right)^{1/2}\), so \(\phi\) is a bounded, hence continuous, linear functional. \(H = \phi^{-1}(\{1\})\) is the preimage of a closed set, so it is closed.H is a translate of the hyperplane \(\ker\phi\): it is not a subspace (it does not contain 0), it is not bounded (add any multiple of a vector in \(\ker\phi\)), and a proper closed hyperplane has no interior points.

Q: Let \(V\) be a closed subspace of \(L^{2}[0,1]\) and let \(f, g \in L^{2}[0,1]\) be given by \(f(x) = x\) and \(g(x) = x^{2}\). If \(V^{\perp} = \text{Span}\{f\}\) and \(Pg\) is the orthogonal projection of \(g\) on \(V\), then \((g - Pg)(x),\ x \in [0,1]\), is

\((g - Pg)(x)\) is \(\frac{3}{4}x\).\(g - Pg\) is the component of \(g\) orthogonal to V, which is the projection of \(g\) onto \(V^{\perp} = \text{Span}\{f\}\):\[ \begin{aligned} g - Pg &= \frac{\langle g, f \rangle}{\langle f, f \rangle} f \\ \langle g, f \rangle &= \int_{0}^{1} x^{3}\,dx = \tfrac{1}{4} \\ \langle f, f \rangle &= \int_{0}^{1} x^{2}\,dx = \tfrac{1}{3} \end{aligned} \]So \(g - Pg = \frac{3}{4}x\). \(\frac{1}{4}x\) forgets to divide by \(\langle f, f \rangle\), and the \(x^{2}\) options are multiples of \(g\) rather than of \(f\).

Q: Let \(\{e_{k} : k \in \mathbb{N}\}\) be an orthonormal basis for a Hilbert space H.Define \(f_{k} = e_{k} + e_{k+1}\), \(k \in \mathbb{N}\) and \(g_{j} = \sum_{n=1}^{j} (-1)^{n+1} e_{n}\), \(j \in \mathbb{N}\).Then \(\sum_{k=1}^{\infty} |\langle g_{j}, f_{k} \rangle|^{2} =\)

The sum equals 1.Write \(g_{j} = \sum c_{n}e_{n}\) with \(c_{n} = (-1)^{n+1}\) for n ≤ j and \(c_{n} = 0\) for n \(\gt\) j. By orthonormality,\[ \langle g_{j}, f_{k} \rangle = c_{k} + c_{k+1} \]For k ≤ j - 1 both coefficients are present with opposite signs, so the inner product is 0. For k = j only \(c_{j} = \pm 1\) survives, giving modulus 1. For k \(\gt\) j both are 0. So the series has a single non-zero term, \(|\pm 1|^{2} = 1\).The options \(j^{2}\) and \(4j^{2}\) come from adding the coefficients without the cancellation between neighbouring terms, and 0 misses the last term k = j.

Q: Let\[ \begin{aligned} \ell^{2} = \{&(x_{1},x_{2},x_{3},\ldots) : x_{n} \in \mathbb{R} \text{ for all } n \in \mathbb{N} \\ &\text{and } \sum_{n=1}^{\infty}x_{n}^{2} \lt \infty\}. \end{aligned} \]For a sequence \((x_{1},x_{2},x_{3},\ldots) \in \ell^{2}\), define\[ \begin{aligned} \|(x_{1},x_{2},x_{3},\ldots)\|_{2} \\ = \left(\sum_{n=1}^{\infty}x_{n}^{2}\right)^{1/2}. \end{aligned} \]Let \(S : (\ell^{2}, \|\cdot\|_{2}) \to (\ell^{2}, \|\cdot\|_{2})\) and \(T : (\ell^{2}, \|\cdot\|_{2}) \to (\ell^{2}, \|\cdot\|_{2})\) be defined by\(S(x_{1},x_{2},x_{3},\ldots) = (y_{1},y_{2},y_{3},\ldots)\), where\[ y_{n} = \begin{cases} 0, & n = 1 \\ x_{n-1}, & n \ge 2 \end{cases} \]\(T(x_{1},x_{2},x_{3},\ldots) = (y_{1},y_{2},y_{3},\ldots)\), where\[ y_{n} = \begin{cases} 0, & n \text{ is odd} \\ x_{n}, & n \text{ is even} \end{cases} \]Then

NEITHER S NOR T is a compact linear map.A compact operator maps the bounded sequence of unit vectors \(e_{n}\) to a sequence with a convergent subsequence.S is the right shift: \(Se_{n} = e_{n+1}\), and \(\|e_{n+1} - e_{m+1}\| = \sqrt{2}\) for \(n \ne m\), so no subsequence converges. S is an isometry on an infinite-dimensional space, so it is not compact.T is the projection onto the even coordinates: \(Te_{2n} = e_{2n}\), again a sequence with mutual distances \(\sqrt{2}\). T is the identity on an infinite-dimensional subspace, so it is not compact either.

Q: Let X be a separable inner product space and Y be any orthonormal set in X. Then :

Y is countable.Distinct elements of an orthonormal set are at distance \(\sqrt{2}\): \(\|e - f\|^{2} = \|e\|^{2} + \|f\|^{2} = 2\). The open balls of radius \(\sqrt{2}/2\) around them are therefore disjoint. A separable space has a countable dense set, and each of these disjoint balls must contain a point of it, so there can be only countably many of them.An orthonormal set is always linearly independent, so A is false. It need not be infinite: in a finite dimensional space it has at most \(n\) elements, so B fails. And in an infinite dimensional space an orthonormal basis is not a Hamel basis, since not every vector is a finite combination of its elements (C).

Q: An infinite dimensional, separable complex Hilbert space is congruent to :

It is congruent to \(l_{2}\).An infinite dimensional separable Hilbert space has a countable orthonormal basis \(\{e_{n}\}\), obtained by Gram-Schmidt from a countable dense set. The map \(x \mapsto (\langle x, e_{n}\rangle)_{n}\) sends it onto the square-summable sequences, and Parseval's identity \(\|x\|^{2} = \sum |\langle x, e_{n}\rangle|^{2}\) shows that the map preserves the inner product. So the space is isometrically isomorphic to \(l_{2}\).\(l_{1}\), \(l_{3}\) and \(l_{\infty}\) are Banach spaces whose norms do not come from an inner product (they fail the parallelogram law), so no Hilbert space can be congruent to them.

Q: Let\[ V = \{f \in C[0, 1] : f(t) = 0\ \forall\ t \le \tfrac12\} \]and\[ W = \{f \in C[0, 1] : f(t) = 0\ \forall\ t \gt \tfrac12\}. \]If C[0, 1] is assigned the inner product\[ \langle f, g\rangle = \int_{0}^{1} f(t)\overline{g(t)}\,dt, \]then :

C[0, 1] is not the sum of V and W.Every \(g \in W\) is continuous and vanishes on \((1/2, 1]\), so \(g(1/2) = 0\) by continuity; every \(f \in V\) vanishes at \(1/2\) by definition. So every element \(f + g\) of \(V + W\) is 0 at \(t = 1/2\).The constant function 1 is in C[0, 1] but is not 0 at 1/2, so \(V + W \ne C[0, 1]\).Since even the sum fails, the space is not a direct sum, orthogonal or otherwise. (V and W are orthogonal to each other, since \(fg = 0\) everywhere, which is what makes options A and B tempting.)

Q: Let X be the space of all continuous complex valued functions on [0, 1] with inner product\[ \langle f, g \rangle = \int_{0}^{1} f(t)\overline{g(t)} dt. \]Let \(T : X \to X\) be defined by \(T(f)(t) = t\, e^{it} f(t)\). Then :

T is normal, but neither self-adjoint nor unitary.T is multiplication by \(\varphi(t) = te^{it}\). Its adjoint is multiplication by \(\overline{\varphi(t)} = te^{-it}\), since \(\langle \varphi f, g \rangle = \int \varphi f \bar{g}\), which equals \(\langle f, \bar{\varphi} g \rangle\). Multiplication operators commute, so \(TT^{*} = T^{*}T\) (both are multiplication by \(t^{2}\)) and T is normal.T is not self-adjoint, because \(\varphi\) is not real valued (\(\varphi \neq \bar{\varphi}\)). It is not unitary, because \(TT^{*}\) is multiplication by \(|\varphi|^{2} = t^{2}\), not the identity. So A, B and D are wrong.

Let \(H\) be a Hilbert space and \(S^{\perp}\) denote the orthogonal complement of a set \(S \subseteq H\). Which of the following is INCORRECT?

A.For \(S_{1}, S_{2} \subseteq H\); \(S_{1} \subseteq S_{2} \Rightarrow S_{1}^{\perp} \subseteq S_{2}^{\perp}\)
B.\(S \subseteq (S^{\perp})^{\perp}\)
C.\(\{0\}^{\perp} = H\)
D.\(S^{\perp}\) is always closed.

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

The incorrect statement is \(S_{1} \subseteq S_{2} \Rightarrow S_{1}^{\perp} \subseteq S_{2}^{\perp}\).

Taking orthogonal complements reverses inclusion. A vector orthogonal to every element of the larger set \(S_{2}\) is orthogonal to every element of \(S_{1}\), so the true statement is \(S_{2}^{\perp} \subseteq S_{1}^{\perp}\). For example, \(\{0\} \subseteq H\) but \(\{0\}^{\perp} = H\) is not contained in \(H^{\perp} = \{0\}\).

The other three are correct. Every \(s \in S\) is orthogonal to every element of \(S^{\perp}\), so \(S \subseteq (S^{\perp})^{\perp}\). Every vector is orthogonal to 0, so \(\{0\}^{\perp} = H\). And \(S^{\perp}\) is an intersection of kernels of the continuous functionals \(x \mapsto \langle x, s \rangle\), hence always closed.

Discuss in Forum

Let \(H = \{(x_{n}) \in \ell_{2} :\) \(\sum_{n=1}^{\infty} \frac{x_{n}}{n} = 1\}\). Then \(H\)

A.is bounded
B.is closed
C.is a subspace
D.has an interior point

Exam Year(s)

GATE Mathematics - 2015,

Answer: B

\(H\) is closed.

Define \(\phi(x) = \sum \frac{x_{n}}{n}\) on \(\ell_{2}\). Since \((\frac{1}{n}) \in \ell_{2}\), the Cauchy-Schwarz inequality gives \(|\phi(x)| \le \|x\|_{2} \left(\sum \frac{1}{n^{2}}\right)^{1/2}\), so \(\phi\) is a bounded, hence continuous, linear functional. \(H = \phi^{-1}(\{1\})\) is the preimage of a closed set, so it is closed.

H is a translate of the hyperplane \(\ker\phi\): it is not a subspace (it does not contain 0), it is not bounded (add any multiple of a vector in \(\ker\phi\)), and a proper closed hyperplane has no interior points.

Discuss in Forum

Let \(V\) be a closed subspace of \(L^{2}[0,1]\) and let \(f, g \in L^{2}[0,1]\) be given by \(f(x) = x\) and \(g(x) = x^{2}\). If \(V^{\perp} = \text{Span}\{f\}\) and \(Pg\) is the orthogonal projection of \(g\) on \(V\), then \((g - Pg)(x),\ x \in [0,1]\), is

A.\(\frac{3}{4}x\)
B.\(\frac{1}{4}x\)
C.\(\frac{3}{4}x^{2}\)
D.\(\frac{1}{4}x^{2}\)

Exam Year(s)

GATE Mathematics - 2015,

Answer: A

\((g - Pg)(x)\) is \(\frac{3}{4}x\).

\(g - Pg\) is the component of \(g\) orthogonal to V, which is the projection of \(g\) onto \(V^{\perp} = \text{Span}\{f\}\):

\[ \begin{aligned} g - Pg &= \frac{\langle g, f \rangle}{\langle f, f \rangle} f \\ \langle g, f \rangle &= \int_{0}^{1} x^{3}\,dx = \tfrac{1}{4} \\ \langle f, f \rangle &= \int_{0}^{1} x^{2}\,dx = \tfrac{1}{3} \end{aligned} \]

So \(g - Pg = \frac{3}{4}x\). \(\frac{1}{4}x\) forgets to divide by \(\langle f, f \rangle\), and the \(x^{2}\) options are multiples of \(g\) rather than of \(f\).

Discuss in Forum

Let \(\{e_{k} : k \in \mathbb{N}\}\) be an orthonormal basis for a Hilbert space H.

Define \(f_{k} = e_{k} + e_{k+1}\), \(k \in \mathbb{N}\) and \(g_{j} = \sum_{n=1}^{j} (-1)^{n+1} e_{n}\), \(j \in \mathbb{N}\).

Then \(\sum_{k=1}^{\infty} |\langle g_{j}, f_{k} \rangle|^{2} =\)

A.0
B.\(j^{2}\)
C.\(4j^{2}\)
D.1

Exam Year(s)

GATE Mathematics - 2022,

Answer: D

The sum equals 1.

Write \(g_{j} = \sum c_{n}e_{n}\) with \(c_{n} = (-1)^{n+1}\) for n ≤ j and \(c_{n} = 0\) for n \(\gt\) j. By orthonormality,

\[ \langle g_{j}, f_{k} \rangle = c_{k} + c_{k+1} \]

For k ≤ j - 1 both coefficients are present with opposite signs, so the inner product is 0. For k = j only \(c_{j} = \pm 1\) survives, giving modulus 1. For k \(\gt\) j both are 0. So the series has a single non-zero term, \(|\pm 1|^{2} = 1\).

The options \(j^{2}\) and \(4j^{2}\) come from adding the coefficients without the cancellation between neighbouring terms, and 0 misses the last term k = j.

Discuss in Forum

Let

\[ \begin{aligned} \ell^{2} = \{&(x_{1},x_{2},x_{3},\ldots) : x_{n} \in \mathbb{R} \text{ for all } n \in \mathbb{N} \\ &\text{and } \sum_{n=1}^{\infty}x_{n}^{2} \lt \infty\}. \end{aligned} \]

For a sequence \((x_{1},x_{2},x_{3},\ldots) \in \ell^{2}\), define

\[ \begin{aligned} \|(x_{1},x_{2},x_{3},\ldots)\|_{2} \\ = \left(\sum_{n=1}^{\infty}x_{n}^{2}\right)^{1/2}. \end{aligned} \]

Let \(S : (\ell^{2}, \|\cdot\|_{2}) \to (\ell^{2}, \|\cdot\|_{2})\) and \(T : (\ell^{2}, \|\cdot\|_{2}) \to (\ell^{2}, \|\cdot\|_{2})\) be defined by

\(S(x_{1},x_{2},x_{3},\ldots) = (y_{1},y_{2},y_{3},\ldots)\), where

\[ y_{n} = \begin{cases} 0, & n = 1 \\ x_{n-1}, & n \ge 2 \end{cases} \]

\(T(x_{1},x_{2},x_{3},\ldots) = (y_{1},y_{2},y_{3},\ldots)\), where

\[ y_{n} = \begin{cases} 0, & n \text{ is odd} \\ x_{n}, & n \text{ is even} \end{cases} \]

Then

A.S is a compact linear map and T is NOT a compact linear map
B.S is NOT a compact linear map and T is a compact linear map
C.both S and T are compact linear maps
D.NEITHER S NOR T is a compact linear map

Exam Year(s)

GATE Mathematics - 2023,

Answer: D

NEITHER S NOR T is a compact linear map.

A compact operator maps the bounded sequence of unit vectors \(e_{n}\) to a sequence with a convergent subsequence.

S is the right shift: \(Se_{n} = e_{n+1}\), and \(\|e_{n+1} - e_{m+1}\| = \sqrt{2}\) for \(n \ne m\), so no subsequence converges. S is an isometry on an infinite-dimensional space, so it is not compact.

T is the projection onto the even coordinates: \(Te_{2n} = e_{2n}\), again a sequence with mutual distances \(\sqrt{2}\). T is the identity on an infinite-dimensional subspace, so it is not compact either.

Discuss in Forum

Let X be a separable inner product space and Y be any orthonormal set in X. Then :

A.Y is linearly dependent
B.Y is countably infinite
C.Y is a basis (Hamel) of X
D.Y is countable

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

Y is countable.

Distinct elements of an orthonormal set are at distance \(\sqrt{2}\): \(\|e - f\|^{2} = \|e\|^{2} + \|f\|^{2} = 2\). The open balls of radius \(\sqrt{2}/2\) around them are therefore disjoint. A separable space has a countable dense set, and each of these disjoint balls must contain a point of it, so there can be only countably many of them.

An orthonormal set is always linearly independent, so A is false. It need not be infinite: in a finite dimensional space it has at most \(n\) elements, so B fails. And in an infinite dimensional space an orthonormal basis is not a Hamel basis, since not every vector is a finite combination of its elements (C).

Discuss in Forum

An infinite dimensional, separable complex Hilbert space is congruent to :

A.\(l_{\infty}\)
B.\(l_{2}\)
C.\(l_{1}\)
D.\(l_{3}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

It is congruent to \(l_{2}\).

An infinite dimensional separable Hilbert space has a countable orthonormal basis \(\{e_{n}\}\), obtained by Gram-Schmidt from a countable dense set. The map \(x \mapsto (\langle x, e_{n}\rangle)_{n}\) sends it onto the square-summable sequences, and Parseval's identity \(\|x\|^{2} = \sum |\langle x, e_{n}\rangle|^{2}\) shows that the map preserves the inner product. So the space is isometrically isomorphic to \(l_{2}\).

\(l_{1}\), \(l_{3}\) and \(l_{\infty}\) are Banach spaces whose norms do not come from an inner product (they fail the parallelogram law), so no Hilbert space can be congruent to them.

Discuss in Forum

Let

\[ V = \{f \in C[0, 1] : f(t) = 0\ \forall\ t \le \tfrac12\} \]

and

\[ W = \{f \in C[0, 1] : f(t) = 0\ \forall\ t \gt \tfrac12\}. \]

If C[0, 1] is assigned the inner product

\[ \langle f, g\rangle = \int_{0}^{1} f(t)\overline{g(t)}\,dt, \]

then :

A.C[0, 1] is the orthogonal direct sum of V and W
B.C[0, 1] is the direct sum of V and W, but not the orthogonal direct sum
C.C[0, 1] is a sum of V and W, but not a direct sum
D.C[0, 1] is not the sum of V and W

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

C[0, 1] is not the sum of V and W.

Every \(g \in W\) is continuous and vanishes on \((1/2, 1]\), so \(g(1/2) = 0\) by continuity; every \(f \in V\) vanishes at \(1/2\) by definition. So every element \(f + g\) of \(V + W\) is 0 at \(t = 1/2\).

The constant function 1 is in C[0, 1] but is not 0 at 1/2, so \(V + W \ne C[0, 1]\).

Since even the sum fails, the space is not a direct sum, orthogonal or otherwise. (V and W are orthogonal to each other, since \(fg = 0\) everywhere, which is what makes options A and B tempting.)

Discuss in Forum

Let \(V_{k} = \{(a_{n}) \in l^{2} : a_{n} = 0\) for all \(n \gt k\}\). Then :

A.\(V_{k}\) is finite dimensional for all \(k\) and \(l^{2}\) is a direct sum of \(\{V_{k} \mid k \in \mathbf{N}\}\)
B.\(V_{k}\) is finite dimensional for all \(k\) and \(l^{2}\) is a sum of \(\{V_{k} \mid k \in \mathbf{N}\}\) and is not a direct sum
C.\(V_{k}\) is infinite dimensional for all \(k\) and \(l^{2}\) is a sum of \(\{V_{k} \mid k \in \mathbf{N}\}\)
D.\(V_{k}\) is finite dimensional for all \(k\) and \(l^{2}\) is not a sum of \(\{V_{k} \mid k \in \mathbf{N}\}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

\(V_{k}\) is finite dimensional for all \(k\) and \(l^{2}\) is not a sum of \(\{V_{k}\}\).

\(V_{k}\) consists of sequences that vanish after the \(k\)-th term, so it is isomorphic to \(\mathbf{R}^{k}\) (or \(\mathbf{C}^{k}\)) and has dimension \(k\). The algebraic sum of the \(V_{k}\) is their union, the sequences with only finitely many non-zero terms.

That is a proper subspace of \(l^{2}\): \((1, \frac{1}{2}, \frac{1}{3}, \ldots)\) is square summable but has infinitely many non-zero terms. So \(l^{2}\) is not a sum (or direct sum) of the \(V_{k}\), only the closure of their union. A and B fail for that reason, and C because each \(V_{k}\) is finite dimensional.

Discuss in Forum

10.

Let X be the space of all continuous complex valued functions on [0, 1] with inner product

\[ \langle f, g \rangle = \int_{0}^{1} f(t)\overline{g(t)} dt. \]

Let \(T : X \to X\) be defined by \(T(f)(t) = t\, e^{it} f(t)\). Then :

A.T is self-adjoint, but not unitary
B.T is unitary, but not self-adjoint
C.T is normal, but neither self-adjoint nor unitary
D.T is not normal

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

T is normal, but neither self-adjoint nor unitary.

T is multiplication by \(\varphi(t) = te^{it}\). Its adjoint is multiplication by \(\overline{\varphi(t)} = te^{-it}\), since \(\langle \varphi f, g \rangle = \int \varphi f \bar{g}\), which equals \(\langle f, \bar{\varphi} g \rangle\). Multiplication operators commute, so \(TT^{*} = T^{*}T\) (both are multiplication by \(t^{2}\)) and T is normal.

T is not self-adjoint, because \(\varphi\) is not real valued (\(\varphi \neq \bar{\varphi}\)). It is not unitary, because \(TT^{*}\) is multiplication by \(|\varphi|^{2} = t^{2}\), not the identity. So A, B and D are wrong.

Inner Product and Hilbert Spaces Questions and Answers

Topics

Other Subjects

Tags: Inner Product and Hilbert Spaces Questions and Answers

About Us

Inner Product and Hilbert Spaces Questions and Answers

Topics

Other Subjects

Tags: Inner Product and Hilbert Spaces Questions and Answers

About Us

Stay informed about all important Dates & Notifications