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Linear Algebra : Linear Mapping

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Example Questions

Example Question #31 : Linear Mapping

A linear map $f:R^4 \rightarrow R^3$ has a rank of 3. Is the linear map, , 1-to-1?

(Hint- Use the formula d=n+r where

is the dimension of the domain

is the dimension of the null space

is the rank of the linear map )

Possible Answers:

Yes

Not enough information

Correct answer:

Explanation:

The answer is no because the dimension of the null space is not zero. This comes from the equation

d=n+r

We know that the domain is R^4 which has dimension of . Therefore

d=4

Also from the problem statement r=2 .

Plugging these into the equation gives

d=n+r

4=n+2

n=2

Since the dimension of the null space is and not , then the function can't be 1-to-1.

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Example Question #31 : Linear Mapping

A linear map $f:R^4 \rightarrow R^2$ has a null space consisting of only the zero vector. Is 1-to-1?

Possible Answers:

Not enough information

Yes

Correct answer:

Yes

Explanation:

If the dimension of the null space is zero then the linear map is 1-to-1.

For this problem, we are told the null space is only the zero vector. Therefore the null space has dimension . Since the null space has dimension , then is 1-to-1.

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Example Question #33 : Linear Mapping

A linear map $f:R^4 \rightarrow R^5$ has a rank of 4. Is the linear map, , 1-to-1?

(Hint- Use the formula d=n+r where

is the dimension of the domain

is the dimension of the null space

is the rank of the linear map )

Possible Answers:

Yes

Not enough information

Correct answer:

Yes

Explanation:

The answer is yes because the dimension of the domain and the rank are equal. This implies that the dimension of the null space is zero.

This comes from the equation

d=n+r

We know that the domain is R^4 which has dimension of . Therefore

d=4

Also from the problem statement r=4 .

Plugging these into the equation gives

d=n+r

4=n+4

n=0

Since the dimension of the null space is then the function is 1-to-1.

Notice that whenever d=r , then is always zero. Thus whenever d=r , the linear mapping is 1-to-1.

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Example Question #32 : Linear Mapping

The null space (sometimes called the kernal) of a mapping is a subspace in the domain such that all vectors in the null space map to the zero vector.

Consider the mapping $f:R^2\rightarrow R^3$ such that f(x,y) = (5x+5y,x+y,0) .

What is the the null space of ?

Possible Answers:

The zero vector

The space spanned by the vector (1,-1)

R^2

R^3

Correct answer:

The space spanned by the vector (1,-1)

Explanation:

Any vector in the null space satisfies f(x,y) = (0,0,0) .

Therefore we get the following equation:

(5x+5y,x+y,0) = (0,0,0)

Thus x=-y . Hence the null space is any vector in form (a,-a) where is any real number. Therefore, any point on the line y=x gets mapped to the zero vector in R^3

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Example Question #34 : Linear Mapping

This problem deals with the zero map. I.e the map that takes all vectors to the zero vector.

Consider the mapping $f:R^2\rightarrow R^3$ such that f(x,y) = (0,0,0) .

What is the the null space of ?

Possible Answers:

The vector (0,0)

The line y=x

R^2

R^3

Correct answer:

R^2

Explanation:

The zero map takes all vectors to the zero vector. Therefore, the entire domain of the map is the null space. The domain of this map is R^2 . Thus R^2 is the null space.

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Example Question #35 : Linear Mapping

This problem deals with the zero map. I.e the map the takes all vectors to the zero vector.

Consider the mapping $f:R^2\rightarrow R^3$ such that f(x,y) = (0,0,0) .

What is the the rank of ?

Possible Answers:

Correct answer:

Explanation:

The image of the the zero map is the zero vector. A single vector has dimension . Therefore the dimension of the image is zero. Hence the rank is zero.

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Example Question #514 : Operations And Properties

$C_{ \mathbb{R}}$ refers to the set of all functions with domain $\mathbb{R}$ and range a subset of $\mathbb{R}$ .

Define the transformation $T: C_{ \mathbb{R}} \rightarrow \mathbb{R}$ to be

$T(f) = \sum_{j=1}^{100} f'(j)$

True or false: is a linear transformation.

Possible Answers:

True

False

Correct answer:

True

Explanation:

For to be a linear transformation, it must hold that

T(f+g)= T(f)+ T(g)

and

T(cf) = cT(f)

for all f,g in the domain of and and for all scalar .

Let $f, g \in C_{ \mathbb{R}}$ .

$T(f) = \sum_{j=1}^{100} f'(j)$ and $T(g) = \sum_{j=1}^{100} g'(j)$ , so

$T(f)+ T(g)= \sum_{j=1}^{100} f'(j)+ \sum_{j=1}^{100} g'(j)$

By the sum rule for finite sequences,

$T(f)+ T(g)= \sum_{j=1}^{100}[ f'(j) + g'(j)]$

By the derivative sum rule,

$T(f)+ T(g)= \sum_{j=1}^{100}( f '+g')(j)$

$T(f)+ T(g)= \sum_{j=1}^{100}( f +g)'(j)$

T(f)+ T(g)= T(f+g)

The first condition is met.

Let $f \in C_{ \mathbb{R}}$ and be a scalar.

$T(f) = \sum_{j=1}^{100} f'(j)$

$c T(f) = c\sum_{j=1}^{100} f'(j)$

By the scalar product rule for finite sequences,

$c T(f) = \sum_{j=1}^{100}c f'(j)$

By the scalar product rule for derivatives,

$c T(f) = \sum_{j=1}^{100}(c f)'(j)$

c T(f) =T(c f)

The second condition is met.

is a linear transformation.

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Example Question #33 : 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:

False

True

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 #34 : 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:

True

False

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 #32 : Linear Mapping

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

Possible Answers:

True

False

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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Linear Algebra : Linear Mapping

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #31 : Linear Mapping

Example Question #31 : Linear Mapping

Example Question #33 : Linear Mapping

Example Question #32 : Linear Mapping

Example Question #34 : Linear Mapping

Example Question #35 : Linear Mapping

Example Question #514 : Operations And Properties

Example Question #33 : Linear Mapping

Example Question #34 : Linear Mapping

Example Question #32 : Linear Mapping

All Linear Algebra Resources

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