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Linear Algebra : Operations and Properties

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Example Questions

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Example Question #31 : Linear Mapping

$C_{[0, \pi]}$ refers to the set of all functions that are continuous on $[0, \pi]$ .

Define the linear mapping $T: C_{[0, \pi]}\rightarrow \mathbb{R}$ as follows:

$T(f)= \int_{0}^{ \pi} f(x) \textup{ d}x$

True or false: $f(x) = \cos x + \sin x$ is in the kernel of .

Possible Answers:

True

False

Correct answer:

False

Explanation:

The kernel of a linear transformation is the subset of the domain of that maps into the zero of its range, so, by definition, $f \in \ker (T)$ if and only if

$T(f)= \int_{0}^{ \pi} f(x) \textup{ d}x = 0$

To determine whether this is true or false, evaluate the integral:

$T(f)= \int_{0}^{ \pi} (\cos x + \sin x) \textup{ d}x = 0$

$=\ \left.\begin{matrix} \sin x - \cos x \\ \; \end{matrix}\right|\ \begin{matrix} \pi \\ 0 \end{matrix}$

$= (\sin \pi - \cos \pi) - (\sin 0 - \cos 0 )$

= (0 - (- 1)) - (0-1)

$T(f) \ne 0$ , so $f \notin \ker (T)$ .

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Example Question #39 : Linear Mapping

$C_{[0, 2\pi]}$ refers to the set of all functions that are continuous on $[0,2\pi]$ .

Define the linear mapping $T: C_{[0, 2\pi]}\rightarrow \mathbb{R}$ as follows:

$T(f)= \int_{0}^{2 \pi} f(x) \textup{ d}x$

True or false: $f(x) = \cos x + \sin x$ is in the kernel of .

Possible Answers:

False

True

Correct answer:

True

Explanation:

The kernel of a linear transformation is the subset of the domain of that maps into the zero of its range. It follows that $f \in \ker (T)$ if and only if

$T(f)= \int_{0}^{2 \pi} f(x) \textup{ d}x = 0$

To determine whether this is true or false, evaluate the integral:

$T(f)= \int_{0}^{2 \pi} (\cos x + \sin x) \textup{ d}x = 0$

$=\ \left.\begin{matrix} \sin x - \cos x \\ \; \end{matrix}\right|\ \begin{matrix} 2 \pi \\ 0 \end{matrix}$

$= (\sin 2\pi - \cos 2 \pi) - (\sin 0 - \cos 0)$

= (0 - 1) - (0-1)

= 0

Therefore, $f \in \ker (T)$ .

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Example Question #521 : Operations And Properties

True or false: If $T: V \rightarrow W$ is a linear mapping, and is a vector space, then is a subspace of .

Possible Answers:

False

True

Correct answer:

False

Explanation:

For example, if is the space of all vectors in $\mathbb{R}^2$ of the form (a,3a) , and is the space of all vectors in $\mathbb{R}^2$ the form (a,0) , then $T: (a_1,a_2) \rightarrow (a_1,0)$ is a linear mapping, but is not a subset of , let alone a subspace of .

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Example Question #521 : Operations And Properties

True or false: The identity mapping $I:V \rightarrow V$ , is also considered a linear mapping, regardless of the vector space .

Possible Answers:

False

True

Correct answer:

True

Explanation:

Verifying the conditions for a linear mapping, we have

$I[a\mathbf{x}+b]=a\mathbf{x}+b = aI[\mathbf{x}]+b.$

Hence the identity mapping is closed under vector addition and scalar multiplication, and is therefore a linear mapping.

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Example Question #42 : Linear Mapping

$P_{N}$ is the set of all polynomials with degree or less.

Define a linear mapping $T: P_{3} \rightarrow P_{4}$ as follows:

$T(p) = \int_{0}^{x}p(y) \textup{ d} y$

Is this mapping one-to-one and onto?

Possible Answers:

No; it is neither.

No; it is onto but not one-to-one.

No; it is one-to-one but not onto.

Yes.

Correct answer:

No; it is one-to-one but not onto.

Explanation:

$P_{N}$ is a vector space of dimension N+1 . $T: P_{3} \rightarrow P_{4}$ is a linear mapping of a four-dimensional vector space into a five-dimensional vector space; since the range has greater dimension than the domain, cannot be onto. It remains to be determined whether it is one-to-one.

A transformation is one-to-one if and only if for any p,q in the domain,

T(p)= T(q ) implies that p = q .

Suppose that T(p)= T(q ) for some $p, q \in P_{3}$ .

Since p,q are third-degree polynomials:

$p(x) = a_{3}x^{3} + a_{2}x^{2}+ a_{1}x+ a_{0}$

$q(x) = b_{3}x^{3} + b_{2}x^{2}+ b_{1}x+ b_{0}$

for some scalar $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

T(p)= T(q ) , so

$\int_{0}^{x}p(y) \textup{ d} y = \int_{0}^{x}q(y) \textup{ d} y$

$\int_{0}^{x} (a_{3}y^{3} + a_{2}y^{2}+ a_{1}y+ a_{0}) \textup{ d} y = \int_{0}^{x} (b_{3}y^{3} + b_{2}y^{2}+ b_{1}y+ b_{0}) \textup{ d} y$

$\frac{a_{3}}{4}x^{4} + \frac{a_{2}}{3}x^{3}+ \frac{a_{1}}{2}x^{2}+ a_{0} x = \frac{b_{3}}{4}x^{4} + \frac{b_{2}}{3}x^{3}+ \frac{b_{1}}{2}x^{2}+ b_{0}x$

For these two polynomials to be equal, it must hold that all coefficients of the terms of equal degree are equal. We can immediately see from the first-degree terms that $a_{0} = b_{0}$ . Applying some simple algebra, it follows almost as quickly that $a_{1} = b_{1}$ , $a_{2} = b_{2}$ , and $a_{3} = b_{3}$ . Thus, p = q , and is one-to-one.

The correct response is that is one-to-one but not onto.

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Example Question #43 : Linear Mapping

$P_{N}$ is the set of all polynomials with degree or less.

Define a linear mapping $T: P_{3} \rightarrow P_{3}$ as follows:

$T(p) = \frac{1}{x}\int_{0}^{x}p(y) \textup{ d} y$

Is this mapping one-to-one and onto?

Possible Answers:

No; it is neither.

Yes.

No; it is onto but not one-to-one.

No; it is one-to-one but not onto.

Correct answer:

Yes.

Explanation:

$T: P_{3} \rightarrow P_{3}$ is a linear mapping of a vector space into itself, so it is possible for to be both one-to-one and onto.

A transformation is one-to-one if and only if for any p,q in the domain,

T(p)= T(q ) implies that p = q .

Suppose that T(p)= T(q ) for some $p, q \in P_{3}$ .

Since p,q are third-degree polynomials:

$p(x) = a_{3}x^{3} + a_{2}x^{2}+ a_{1}x+ a_{0}$

$q(x) = b_{3}x^{3} + b_{2}x^{2}+ b_{1}x+ b_{0}$

for some scalar $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

T(p)= T(q ) , so

$\frac{1}{x}\int_{0}^{x}p(y) \textup{ d} y = \frac{1}{x}\int_{0}^{x}q(y) \textup{ d} y$

$\int_{0}^{x}p(y) \textup{ d} y = \int_{0}^{x}q(y) \textup{ d} y$

$\int_{0}^{x} (a_{3}y^{3} + a_{2}y^{2}+ a_{1}y+ a_{0}) \textup{ d} y = \int_{0}^{x} (b_{3}y^{3} + b_{2}y^{2}+ b_{1}y+ b_{0}) \textup{ d} y$

$\frac{a_{3}}{4}x^{4} + \frac{a_{2}}{3}x^{3}+ \frac{a_{1}}{2}x^{2}+ a_{0} x = \frac{b_{3}}{4}x^{4} + \frac{b_{2}}{3}x^{3}+ \frac{b_{1}}{2}x^{2}+ b_{0}x$

A transformation is onto if, for each in the range, there exists in the domain such that T(p) = r .

Let $r(x) \in P_{3}$ . Then

$r(x) = c_{3}x^{3} + c_{2}x^{2}+ c_{1}x+ c_{0}$

for some scalar $\left \{ c_{0}, c_{1}, c_{2},c_{3} \right \}$ .

If T(p) = r , then, if p(x) is defined as before,

$\frac{1}{x}\int_{0}^{x}p(y) \textup{ d} y = r(x)$

$\frac{1}{x}\int_{0}^{x} (a_{3}y^{3} + a_{2}y^{2}+ a_{1}y+ a_{0}) \textup{ d} y = c_{3}x^{3} + c_{2}x^{2}+ c_{1}x+ c_{0}$

$\frac{1}{x} \left ( \frac{a_{3}}{4}x^{4} + \frac{a_{2}}{3}x^{3}+ \frac{a_{1}}{2}x^{2}+ a_{0} x \right ) = c_{3}x^{3} + c_{2}x^{2}+ c_{1}x+ c_{0}$

$\ \frac{a_{3}}{4}x^{3} + \frac{a_{2}}{3}x^{2}+ \frac{a_{1}}{2}x + a_{0} = c_{3}x^{3} + c_{2}x^{2}+ c_{1}x+ c_{0}$

Therefore,

$\ \frac{a_{3}}{4}= c_{3}$ , $\frac{a_{2}}{3}= c_{2}$ , $\frac{a_{1}}{2}= c_{1}$ , $a_{0} = c_{0}$

Or,

$a_{3} = 4c_{3}$ , $a_{2} =3 c_{2}$ , $a_{1} =2 c_{1}$ , $a_{0} = c_{0}$ .

Thus,

$p(x) = 4c_{3}x^{3}+ 3c_{2}x^{2}+ 2c_{1}x+ c_{0}$ is the polynomial in $P_{3}$ that maps into r(x) . Since such a polynomial exists in the domain for each range element, it follows that is onto.

The correct response is that is one-to-one and onto.

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Example Question #44 : Linear Mapping

$P_{3}$ is the set of all real polynomials with degree 3 or less.

Define the linear mapping $T: P_{3}\rightarrow \mathbb{R}$ as follows:

$T(P(x)) \rightarrow P(1)$

Is this linear mapping one-to-one and onto?

Possible Answers:

No; it is one-to-one but not onto.

Yes.

No; it is onto but not one-to-one.

No; it is neither.

Correct answer:

No; it is onto but not one-to-one.

Explanation:

$T: P_{3}\rightarrow \mathbb{R}$ is a linear mapping of a four-dimensional vector space into a one-dimensional vector space; it cannot be one-to-one.

is onto if, for every element in its codomain, which here is $\mathbb{R}$ , there exists at least one in the domain $P_{3}$ so that T(x) = y .

For each , we can choose the constant polynomial p(x) = y , so each has at least one domain element that maps into it.

is onto but not one-to-one.

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Example Question #45 : Linear Mapping

$P_{3}$ is the set of all polynomials with degree 3 or less.

Define the linear mapping $T: P_{3}\rightarrow P_{3}$ as follows:

$T( P(x)) \rightarrow xP'(x)$

Is this linear mapping one-to-one and onto?

Possible Answers:

Yes

No; it is one-to-one but not onto.

No; it is neither.

No; it is onto but not one-to-one.

Correct answer:

No; it is neither.

Explanation:

A linear mapping is one-to-one if, for every p, q in the domain such that $p \ne q$ , it must follow that $T(p) \ne T(q)$ . We can show that is not one-to-one by finding $p, q \in P_{3}$ such that T(p) = T(q) . For example, let

p (x) = x+ 3

q(x) = x+ 4

Then

T(p(x)) = xp'(x) = x(1) = x

T(q(x)) = xq'(x) = x(1) = x

Since there exists $p \ne q$ such that T(p) = T(q) , is not one-to-one.

is onto if, for every element in its codomain, which here is $P_{3}$ , there exists at least one in the domain $P_{3}$ so that T(p) = r .

Suppose $r = x^{3} + 1$ . To find such that T(p) =r , set:

$x p'(x)= x^{3}+1$

$p'(x)= x^{2}+ \frac{1}{x}$

$p(x) = \frac{1}{3} x^{3} + \ln x + C$ for some constant .

However, $p \notin P_{3}$ .

Therefore, there does not exist $p \in P_{3}$ such that T(p) = r .

The correct response is that is not one-to-one or onto.

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Example Question #46 : Linear Mapping

$P_{N}$ is the set of all polynomials with degree or less;

Define a transformation $T: P_{3} \rightarrow P_{4}$ as follows:

$T(p) = \int_{1}^{x}p(y) \textup{ d} y$

True or false: is an example of a linear mapping.

Possible Answers:

True

False

Correct answer:

True

Explanation:

is a linear mapping if and only if the following two conditions hold:

Additivity: T(p+q) = T(p)+ T(q) for all $p,q\in P_{3}$ ,

Homogeneity: T(cp ) =cT(p) for all $p \in P_{3} , c$ scalar.

We know from calculus that both properties hold for any definite integrals, so

T(p+q)

$= \int_{1}^{x}(p+q) (y) \textup{ d} y$

$= \int_{1}^{x} p(y) \textup{ d} y + \int_{1}^{x} q(y) \textup{ d} y$

= T(p)+ T(q)

and

T(cp )

$= \int_{1}^{x} (cp)(y) \textup{ d} y$

$= c\int_{1}^{x} p(y) \textup{ d} y$

=c T(p)

This makes a linear mapping.

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Example Question #521 : Operations And Properties

$P_{3}$ is the set of all polynomials with degree 3 or less.

Define a transformation $T: P_{3} \rightarrow P_{3}$ as follows:

$T( p) = x^{2}p''$ .

True or false: is an example of a linear mapping.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is a linear mapping if and only if the following two conditions hold:

Additivity: T(p+q) = T(p)+ T(q) for all $p, q \in P_{3}$ ,

Homogeneity: T(cp ) =cT(p) for all $p \in P_{3} , c$ scalar.

Let $p, q \in P_{3}$ . Then

$T( p(x)) = x^{2}p''(x)$

$T( q(x)) = x^{2}q''(x)$

$T( p(x)) + T( q(x)) = x^{2}p''(x) + x^{2}q''(x)$

By distribution:

$T( p(x)) + T( q(x)) = x^{2}(p''(x) + q''(x))$

$T( p(x)) + T( q(x)) = x^{2}((p'' + q'')(x))$

By the sum rule of derivatives:

$x^{2}((p'' + q'')(x)) = x^{2}((p + q )''(x))=T((p+q )(x))$

Thus,

T(p+q) = T(p)+ T(q) ,

proving that additivity holds.

Let $p \in P_{3}$ and be a scalar. Then

$T( p(x)) = x^{2}p''(x)$

$cT( p(x)) =c x^{2}p''(x)$

$cT( p(x)) = x^{2}(cp''(x))$

By the scalar product rule of derivatives,

$x^{2}(cp''(x)) = x^{2} ((cp)''(x)) = T((cp)(x))$ ,

and

T(cp ) =cT(p) ,

proving that homogeneity holds.

is a linear mapping.

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Linear Algebra : Operations and Properties

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #31 : Linear Mapping

Example Question #39 : Linear Mapping

Example Question #521 : Operations And Properties

Example Question #521 : Operations And Properties

Example Question #42 : Linear Mapping

Example Question #43 : Linear Mapping

Example Question #44 : Linear Mapping

Example Question #45 : Linear Mapping

Example Question #46 : Linear Mapping

Example Question #521 : Operations And Properties

All Linear Algebra Resources

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