Linear Mapping - Linear Algebra

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Question

Consider the mapping $f:R^2\rightarrow R^3$ such that .

What is the the null space of ?

Answer

To find the null space consider the equation

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

Thus the null space consists of the single vector

Compare your answer with the correct one above

Question

$P_{N}$ 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?

Answer

$P_{N}$ is a vector space of dimension . $T: P_{3} \rightarrow P_{4}$ 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 in the domain,

implies that .

Suppose that for some $p, q \in P_{3}$ .

Since 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 $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

, 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, , and is one-to-one.

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

Compare your answer with the correct one above

Question

$M_{2 \times 1 }$ is the set of all two-by-one matrices - that is, the set of all column matrices with two entries.

Let $X= \begin{bmatrix} \sqrt{6} & -2\sqrt{3} \- 2\sqrt{3} & 2 \sqrt{6}\end{bmatrix}$ . Define a linear mapping $T: M_{2 \times 1 } \rightarrow M_{2 \times 1 }$ as follows:

.

True or false: is one-to-one and onto.

Answer

The domain and the codomain of are identical, so is one to one if and only if it is onto.

A necessary and sufficient condition for to be one-to-one is that the kernel of be $\left \{ 0 \right \}$ . In $M_{2 \times 1 }$ , the zero element is $O= \begin{bmatrix} 0 \ 0 \end{bmatrix}$ , and this condition states that if

, then

Thus, we can prove that is not one-to-one - and not onto - by finding a nonzero column matrix such that .

Set $A= \begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix}$ . Then

$= \begin{bmatrix} \sqrt{6} & -2\sqrt{3} \- 2\sqrt{3} & 2 \sqrt{6}\end{bmatrix}\begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix}$

$= \begin{bmatrix} \sqrt{6}( \sqrt{2} )+( -2\sqrt{3}) (1)\-2 \sqrt{3} (\sqrt{2}) +( 2\sqrt{6}) (1) \end{bmatrix}$

$= \begin{bmatrix} 2\sqrt{3} -2\sqrt{3} \-2 \sqrt{6}+ 2\sqrt{6} \end{bmatrix}$

$=\begin{bmatrix} 0 \ 0 \end{bmatrix} = O$

There is at least one nonzero column matrix in the kernel of , so is not one-to-one or onto.

Compare your answer with the correct one above

Question

Consider the mapping $f:R^2\rightarrow R^3$ such that .

What is the the null space of ?

Answer

To find the null space consider the equation

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

Thus the null space consists of the single vector

Compare your answer with the correct one above

Question

$P_{N}$ 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?

Answer

$P_{N}$ is a vector space of dimension . $T: P_{3} \rightarrow P_{4}$ 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 in the domain,

implies that .

Suppose that for some $p, q \in P_{3}$ .

Since 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 $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

, 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, , and is one-to-one.

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

Compare your answer with the correct one above

Question

$P_{N}$ 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?

Answer

$T: P_{3} \rightarrow P_{3}$ 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 in the domain,

implies that .

Suppose that for some $p, q \in P_{3}$ .

Since 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 $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

, so

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

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

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

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

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, , and is one-to-one.

A transformation is onto if, for each in the range, there exists in the domain such that .

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 , then, if 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 . 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.

Compare your answer with the correct one above

Question

$M_{2 \times 1 }$ is the set of all two-by-one matrices - that is, the set of all column matrices with two entries.

Let $X= \begin{bmatrix} \sqrt{6} & -2\sqrt{3} \- 2\sqrt{3} & 2 \sqrt{6}\end{bmatrix}$ . Define a linear mapping $T: M_{2 \times 1 } \rightarrow M_{2 \times 1 }$ as follows:

.

True or false: is one-to-one and onto.

Answer

The domain and the codomain of are identical, so is one to one if and only if it is onto.

A necessary and sufficient condition for to be one-to-one is that the kernel of be $\left \{ 0 \right \}$ . In $M_{2 \times 1 }$ , the zero element is $O= \begin{bmatrix} 0 \ 0 \end{bmatrix}$ , and this condition states that if

, then

Thus, we can prove that is not one-to-one - and not onto - by finding a nonzero column matrix such that .

Set $A= \begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix}$ . Then

$= \begin{bmatrix} \sqrt{6} & -2\sqrt{3} \- 2\sqrt{3} & 2 \sqrt{6}\end{bmatrix}\begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix}$

$= \begin{bmatrix} \sqrt{6}( \sqrt{2} )+( -2\sqrt{3}) (1)\-2 \sqrt{3} (\sqrt{2}) +( 2\sqrt{6}) (1) \end{bmatrix}$

$= \begin{bmatrix} 2\sqrt{3} -2\sqrt{3} \-2 \sqrt{6}+ 2\sqrt{6} \end{bmatrix}$

$=\begin{bmatrix} 0 \ 0 \end{bmatrix} = O$

There is at least one nonzero column matrix in the kernel of , so is not one-to-one or onto.

Compare your answer with the correct one above

Question

Consider the mapping $f:R^2\rightarrow R^3$ such that .

What is the the null space of ?

Answer

To find the null space consider the equation

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

Thus the null space consists of the single vector

Compare your answer with the correct one above

Question

$P_{N}$ 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?

Answer

$P_{N}$ is a vector space of dimension . $T: P_{3} \rightarrow P_{4}$ 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 in the domain,

implies that .

Suppose that for some $p, q \in P_{3}$ .

Since 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 $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

, 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, , and is one-to-one.

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

Compare your answer with the correct one above

Question

$P_{N}$ 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?

Answer

$T: P_{3} \rightarrow P_{3}$ 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 in the domain,

implies that .

Suppose that for some $p, q \in P_{3}$ .

Since 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 $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

, so

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

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

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

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

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, , and is one-to-one.

A transformation is onto if, for each in the range, there exists in the domain such that .

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 , then, if 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 . 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.

Compare your answer with the correct one above

Question

$M_{2 \times 1 }$ is the set of all two-by-one matrices - that is, the set of all column matrices with two entries.

Let $X= \begin{bmatrix} \sqrt{6} & -2\sqrt{3} \- 2\sqrt{3} & 2 \sqrt{6}\end{bmatrix}$ . Define a linear mapping $T: M_{2 \times 1 } \rightarrow M_{2 \times 1 }$ as follows:

.

True or false: is one-to-one and onto.

Answer

The domain and the codomain of are identical, so is one to one if and only if it is onto.

A necessary and sufficient condition for to be one-to-one is that the kernel of be $\left \{ 0 \right \}$ . In $M_{2 \times 1 }$ , the zero element is $O= \begin{bmatrix} 0 \ 0 \end{bmatrix}$ , and this condition states that if

, then

Thus, we can prove that is not one-to-one - and not onto - by finding a nonzero column matrix such that .

Set $A= \begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix}$ . Then

$= \begin{bmatrix} \sqrt{6} & -2\sqrt{3} \- 2\sqrt{3} & 2 \sqrt{6}\end{bmatrix}\begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix}$

$= \begin{bmatrix} \sqrt{6}( \sqrt{2} )+( -2\sqrt{3}) (1)\-2 \sqrt{3} (\sqrt{2}) +( 2\sqrt{6}) (1) \end{bmatrix}$

$= \begin{bmatrix} 2\sqrt{3} -2\sqrt{3} \-2 \sqrt{6}+ 2\sqrt{6} \end{bmatrix}$

$=\begin{bmatrix} 0 \ 0 \end{bmatrix} = O$

There is at least one nonzero column matrix in the kernel of , so is not one-to-one or onto.

Compare your answer with the correct one above

Question

Consider the mapping $f:R^2\rightarrow R^3$ such that .

What is the the null space of ?

Answer

To find the null space consider the equation

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

Thus the null space consists of the single vector

Compare your answer with the correct one above

Question

$P_{N}$ 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?

Answer

$P_{N}$ is a vector space of dimension . $T: P_{3} \rightarrow P_{4}$ 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 in the domain,

implies that .

Suppose that for some $p, q \in P_{3}$ .

Since 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 $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

, 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, , and is one-to-one.

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

Compare your answer with the correct one above

Question

$P_{N}$ 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?

Answer

$T: P_{3} \rightarrow P_{3}$ 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 in the domain,

implies that .

Suppose that for some $p, q \in P_{3}$ .

Since 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 $\left \{ a_{0}, ...a_{3}, b_{0}, ...b_{3} \right \}$ .

, so

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

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

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

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

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, , and is one-to-one.

A transformation is onto if, for each in the range, there exists in the domain such that .

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 , then, if 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 . 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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Question

$M_{2 \times 1 }$ is the set of all two-by-one matrices - that is, the set of all column matrices with two entries.

Let $X= \begin{bmatrix} \sqrt{6} & -2\sqrt{3} \- 2\sqrt{3} & 2 \sqrt{6}\end{bmatrix}$ . Define a linear mapping $T: M_{2 \times 1 } \rightarrow M_{2 \times 1 }$ as follows:

.

True or false: is one-to-one and onto.

Answer

The domain and the codomain of are identical, so is one to one if and only if it is onto.

A necessary and sufficient condition for to be one-to-one is that the kernel of be $\left \{ 0 \right \}$ . In $M_{2 \times 1 }$ , the zero element is $O= \begin{bmatrix} 0 \ 0 \end{bmatrix}$ , and this condition states that if

, then

Thus, we can prove that is not one-to-one - and not onto - by finding a nonzero column matrix such that .

Set $A= \begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix}$ . Then

$= \begin{bmatrix} \sqrt{6} & -2\sqrt{3} \- 2\sqrt{3} & 2 \sqrt{6}\end{bmatrix}\begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix}$

$= \begin{bmatrix} \sqrt{6}( \sqrt{2} )+( -2\sqrt{3}) (1)\-2 \sqrt{3} (\sqrt{2}) +( 2\sqrt{6}) (1) \end{bmatrix}$

$= \begin{bmatrix} 2\sqrt{3} -2\sqrt{3} \-2 \sqrt{6}+ 2\sqrt{6} \end{bmatrix}$

$=\begin{bmatrix} 0 \ 0 \end{bmatrix} = O$

There is at least one nonzero column matrix in the kernel of , so is not one-to-one or onto.

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Question

Consider the mapping $f:R^2\rightarrow R^3$ . Can f be an isomorphism?

(Hint: Think about dimension's role in isomorphism)

Answer

No, f, cannot be an isomorphism. This is because and have different dimension. Isomorphisms cannot exist between vector spaces of different dimension.

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Question

The last question showed us isomorphisms must be between vector spaces of the same dimension. This question now asks about homomorphisms.

Consider the mapping $f:R^2\rightarrow R^3$ . Can f be a homomorphism?

Answer

The answer is yes. There is no restriction on dimension for homomorphism like there is for isomorphism. Therefore f could be a homomorphism, but it is not guaranteed.

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Question

In the previous question, we said an isomorphism cannot be between vector spaces of different dimension. But are all homomorphisms between vector spaces of the same dimension an isomorphism?

Consider the homomorphism $f:R^2\rightarrow R^2$ . Is f an isomorphism?

Answer

The answer is not enough information. The reason is that it could be an isomorphism because it is between vector spaces of the same dimension, but that doesn't mean it is.

For example:

Consider the zero mapping f(x,y)= (0,0).

This mapping is not onto or 1-to-1 because all elements go to the zero vector. Therefore it is not an isomorphism even though it is a mapping between spaces with the same dimension.

Another example:

Consider the identity mapping f(x,y) = (x,y)

This is an isomorphism. It clearly preserves structure and is both onto and 1-to-1.

Thus f could be an isomorphism (example identity map) or it could NOT be an isomorphism ( Example the zero mapping)

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Question

Let f be a homomorphism from to . Can f be 1-to-1?

(Hint: look at the dimension of the domain and co-domain)

Answer

No, f can not be 1-to-1. The reason is because the domain has dimension 3 but the co-domain has dimension of 2. A mapping can not be 1-to-1 when the the dimension of the domain is greater than the dimension of the co-domain.

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Question

Often we can get information about a mapping by simply knowing the dimension of the domain and codomain.

Let f be a mapping from to . Can f be onto?

(Hint look at the dimension of the domain and codomain)

Answer

No, f cannot be onto. The reason is because the dimension of the domain (2) is less than the dimension of the codomain(3).

For a function to be onto, the dimension of the domain must be less than or equal to the dimension of the codomain.

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