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Linear Algebra : Linear Mapping

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Example Question #491 : Operations And Properties

Let f be a mapping such that $f:R^2 \rightarrow P^2$ where P^2 is the vector space of polynomials up to the x^2 term. (ie polynomials of the form a_0+a_1 x+ a_2 x^2 )

Let f be defined such that f(a,b) = 2b + (a+b)x + 2ax^2

Is f a homomorphism?

Possible Answers:

Yes

No, because both scalar multiplication and vector addition is not preserved

No, because scalar multiplication is not preserved

No because vector addition is not preserved

Correct answer:

Yes

Explanation:

f is a homomorphism because it preserves both vector addition and scalar multiplication.

To show this we need to prove both statements

Proof f preserves vector addition

Let u and v be arbitrary vectors in R^2 with the form u=(u_1,u_2) and v=(v_1,v_2)

Consider f(u+v)=f((u_1,u2)+(v_1,v_2)) = f(u_1+v_1,u_2+v_2) . Applying the definition of f we get

f(u+v)= 2(u_2+v_2)+(u_1+v_1+u_2+v_2)x+2(u_1+v_1)x^2

This is the same thing as f(u)+f(v) = 2(u_2)+(u_1+u_2)x+2(u_1)x^2 +2(v_2)+(v_1+v_2)x+2(v_1)x^2 = 2(u_2+v_2)+(u_1+v_1+u_2+v_2)x+2(u_1+v_1)x^2

Hence, f preserves vector addition because f(u+v) = f(u)+f(v)

Proof f preserves scalar multiplication

Let u be an arbitrary vector in R^2 with the form u=(u_1,u_2) and let k be an arbitrary real constant.

Consider f(ku)=f(ku_1,ku_2) = 2ku_2+(ku_1+ku_2)x+2(ku_1)x^2

This is the same thing we get if we consider k f(u)=kf(u_1,u_2) = k(2u_2+(u_1+u_2)x+2(u_1)x^2) =f(ku)=f(ku_1,ku_2) = 2ku_2+(ku_1+ku_2)x+2(ku_1)x^2

Hence f preserves scalar multiplication because f(ku)=kf(u) for all vectors u and scalars k.

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Example Question #491 : Operations And Properties

Let f be a mapping such that $f:R^2 \rightarrow P^2$ where P^2 is the vector space of polynomials up to the x^2 term. (ie polynomials of the form a_0+a_1 x+ a_2 x^2 )

Let f be defined such that f(a,b) = 2b + (a+b)x + 2ax^2

Is f an isomorphism?

(Hint: The last problem we showed this particular f is a homomorphism)

Possible Answers:

No, f doesn't preserve vector addition and scalar multiplication

Yes

No, f is not 1-to-1

No, f is not onto

Correct answer:

No, f is not onto

Explanation:

f is not onto. This is because not every vector in P^2 is in the image of f. For example, the vector 1+x+x^2 is not in the image of f. Hence, f is not onto.

We could also see this quicker by looking at the dimension of the domain and codomain. The domain R^2 has dimension 2 and the codomain P^2 has dimension 3. A mapping can't be onto and have a domain with a lower dimension than the codomain.

Finally, we know f preserves vector addition and scalar multiplication because it is a homomorphism.

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Example Question #871 : Linear Algebra

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

What is the the null space of ?

Possible Answers:

R^2

The vector (0,0)

R^3

The line x=y

Correct answer:

The vector (0,0)

Explanation:

To find the null space consider the equation

f(x,y) = (0,0,0)

(x-y,x+y,2x+3y) = (0,0,0)

This gives a system of equations

$\left\{\begin{matrix} 0&=x-y \\ 0&=x+y \\ 0&=2x+3y \end{matrix}\right.$

The only solution to this system is x=y=0

Thus the null space consists of the single vector (0,0)

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Example Question #14 : Linear Mapping

An important quantity for a linear map is the dimension of its image. This is called the rank.

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

What is the the rank of ?

Possible Answers:

Correct answer:

Explanation:

To find the rank of , we first find the image of . The image of is any vector of form (5a,a,0) where a is any real number. This is a line in R^3 . Therefore the image of is a 1 dimensional subspace. Thus the answer is 1.

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Example Question #491 : Operations And Properties

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

What is the the rank of ?

Possible Answers:

Correct answer:

Explanation:

The image is the space spanned by the vectors (1,1,2) and (-1,1,3) . The image has a basis of vectors. Therefore the image has dimension . Thus the rank is .

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Example Question #871 : Linear Algebra

The dimension of the domain can be used to learn about the dimension of the null space and the rank of a linear map.

Let be a linear map such that $f:R^2\rightarrow R^3$ . What is the maximum possible rank of .

Possible Answers:

Correct answer:

Explanation:

The maximum possible rank of a function is the dimension of the domain. The domain of is R^2 . Therefore the maximum possible rank of is .

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Example Question #17 : Linear Mapping

The dimension of the domain can be used to learn about the dimension of the null space and the rank of a linear map.

Let be a linear map such that $f:R^2\rightarrow R^3$ . What is the maximum dimension of the null space of ?

Possible Answers:

Correct answer:

Explanation:

The null space is the subspace such that maps to the zero vector in the codomain. The largest possible null space is when the entire domain goes to the zero vector. The domain of is R^2 . Therefore the largest possible null space would be R^2 which has a dimension of . Thus the largest possible dimension for the null space is .

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Example Question #18 : Linear Mapping

There is a relationship between the dimension of the domain and the dimension of the null space and the rank of a function. The relationship is

d=n+r

where is the dimension of the domain,

is the dimension of the null space,

and is the rank.

Let be linear map such that $f:R^5\rightarrow R^9$ and the rank is 3. What is the dimension of the null space of ?

Possible Answers:

Correct answer:

Explanation:

The answer is .

The dimension of the domain is .

The rank of f is .

Plug these into the equation given above to get

5 = n+2

n=3

Therefore, the dimension of the null space is .

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Example Question #19 : Linear Mapping

There is a relationship between the dimension of the domain and the dimension of the null space and the rank of a function. This question is on that.

Let be linear map with a rank of 6 and a null space with dimension . What is the dimension of the domain of ?

Possible Answers:

Correct answer:

Explanation:

Use the formula

d=n+r

where is the dimension of the domain,

is the dimension of the null space,

and is the rank.

From the problem statement, we know n=3 and r=6 . Therefore

d = n + r = 3 + 6 = 9

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Example Question #20 : Linear Mapping

Let be a linear mapping such that $f:R^4\rightarrow R^5$ . What is the largest possible rank of ?

Possible Answers:

Correct answer:

Explanation:

The answer is . The largest possible rank of a linear map is always the dimension of the domain. Let's see why that is.

Use the formula

d=n+r

where is the dimension of the domain,

is the dimension of the null space,

and is the rank.

From the problem statement, we know d=4 and is largest when n=0 . Hence

d = n + r

4= 0 + r

r=4.

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Linear Algebra : Linear Mapping

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #491 : Operations And Properties

Example Question #491 : Operations And Properties

Example Question #871 : Linear Algebra

Example Question #14 : Linear Mapping

Example Question #491 : Operations And Properties

Example Question #871 : Linear Algebra

Example Question #17 : Linear Mapping

Example Question #18 : Linear Mapping

Example Question #19 : Linear Mapping

Example Question #20 : Linear Mapping

All Linear Algebra Resources

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