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Linear Algebra : Operations and Properties

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Example Questions

Example Question #904 : Linear Algebra

$M_{2 \times 2 }$ is the set of all two-by-two matrices.

Define the linear mapping $T:M_{2 \times 2 } \rightarrow M_{2 \times 2 }$ as follows:

$T\left ( \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \right ) = \begin{bmatrix} a_{11} &2 a_{12} \\ 3a_{21} &4 a_{22} \end{bmatrix}$

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

Possible Answers:

False; is neither one-to-one nor onto

False; is onto but not one-to-one

True

False; is one-to-one but not onto

Correct answer:

True

Explanation:

The domain and the codomain of are identical, so is one to one if and only if it is onto. It suffices to test either condition; so it will be determined whether is onto.

is onto if, for each $B \in M_{2\times 2 }$ , there exists $A \in M_{2\times 2 }$ such that T(A)= B . Let

$B = \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}$

Then, if

$A = \begin{bmatrix} b_{11} &\frac{1}{2} b_{12}\\ \\ \frac{1}{3}b_{21} & \frac{1}{4}b_{22} \end{bmatrix}$ ,

then

$T\left ( \begin{bmatrix} b_{11} &\frac{1}{2} b_{12}\\ \\ \frac{1}{3}b_{21} & \frac{1}{4}b_{22} \end{bmatrix} \right ) = \begin{bmatrix} b_{11} &2 \cdot \frac{1}{2} b_{12}\\ \\3 \cdot \frac{1}{3}b_{21} & 4\cdot \frac{1}{4}b_{22} \end{bmatrix}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}= B$

T(A)= B .

is onto; it follows that is also one-to-one.

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Example Question #905 : Linear Algebra

$M_{2 \times 2 }$ is the set of all two-by-two matrices.

Define the mapping $T:M_{2 \times 2 } \rightarrow M_{2 \times 2 }$ as follows:

$T\left ( \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \right ) = \begin{bmatrix} a_{11} &2 a_{12} \\ 3a_{21} &4 a_{22} \end{bmatrix}$

True or false: is a linear mapping.

Possible Answers:

True

False

Correct answer:

True

Explanation:

is a linear mapping if two conditions hold:

Additivity:

For all $A, B \in M_{2 \times 2}$

T(A+B)= T(A)+ T(B)

Homogeneity:

For all $A \in M_{2 \times 2}$ and scalar ,

T(cA) = cT(A)

First, test for additivity.

Let $A, B \in M_{2 \times 2}$

Then

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ and $B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$ ,

and

$A+B = \begin{bmatrix}a_{11} + b_{11} &a_{12}+ b_{12} \\ a_{21}+ b_{21} & a_{22} + b_{22} \end{bmatrix}$ .

T(A+B)

$=T\left ( \begin{bmatrix}a_{11} + b_{11} &a_{12}+ b_{12} \\ a_{21}+ b_{21} & a_{22} + b_{22} \end{bmatrix}\right )$

$= \begin{bmatrix} a_{11} + b_{11} &2 (a_{12}+ b_{12} )\\ 3(a_{21}+ b_{21} )&4 (a_{22} + b_{22} )\end{bmatrix}$

$= \begin{bmatrix} a_{11} + b_{11} &2 a_{12}+2 b_{12} \\ 3a_{21}+ 3b_{21} &4 a_{22} + 4 b_{22} \end{bmatrix}$

$= \begin{bmatrix} a_{11} &2 a_{12} \\ 3a_{21} &4 a_{22} \end{bmatrix}+ \begin{bmatrix} b_{11} &2 b_{12} \\ 3b_{21} &4 b_{22} \end{bmatrix}$

$=T\left ( \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \right ) + T\left ( \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \right )$

=T(A)+ T(B)

Additivity is satisfied.

Now test for homogeneity. Let be a scalar. Then

$cA = \begin{bmatrix} ca_{11} & ca_{12} \\ ca_{21} &c a_{22} \end{bmatrix} \right )$ .

$T(cA) =T \left(\ \begin{bmatrix} ca_{11} & ca_{12} \\ ca_{21} &c a_{22} \end{bmatrix} \right )$

$= \begin{bmatrix} ca_{11} & 2 (ca_{12} )\\ 3 (ca_{21} )&4( c a_{22} )\end{bmatrix}$

$= \begin{bmatrix} ca_{11} & (2 c)a_{12} )\\ (3c) a_{21} )&(4c) a_{22} )\end{bmatrix}$

$= \begin{bmatrix} ca_{11} & (c \cdot 2)a_{12} \\ (c \cdot 3) a_{21} &(c \cdot 4) a_{22} \end{bmatrix}$

$= \begin{bmatrix} ca_{11} & c (2a_{12}) \\ c( a_{21}) &c (4a_{22}) \end{bmatrix}$

$=c \begin{bmatrix} a_{11} &2 a_{12} \\ 3a_{21} &4 a_{22} \end{bmatrix}$

$=c T\left ( \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \right )$

=cT(A)

Homogeneity is satisfied.

is a linear mapping.

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Example Question #906 : Linear Algebra

is the set of all polynomials of finite degree in .

Define mapping $T:P \rightarrow P$ as follows:

T( p(x))= xp(x)+ 1

True or false: is a linear mapping.

Possible Answers:

True

False

Correct answer:

False

Explanation:

is a linear mapping if two conditions hold:

Additivity:

For all $p, q \in P$

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

Homogeneity:

For all $p \in P$ and scalar ,

T(cP) = cT(P)

Homogeneity can be disproved through counterexample.

Let p(x)= x and c= 2 ..

Then

$T(p(x))= x (x)+1 = x^{2}+ 1$ ,

and

$2 T(p(x))=2( x^{2}+ 1) = 2x^{2}+ 2$

However

2p(x)= 2x

$T(2p(x))= x (2x)+1 = 2x^{2}+ 1$

$T(2p(x)) \ne 2T(p(x))$ ,

so homogeneity does not hold in general. is not a linear mapping.

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Example Question #51 : Linear Mapping

$M_{2 \times 1 }$ 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:

T(A) =XA .

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

Possible Answers:

True

False; is neither one-to-one nor onto

False; is one-to-one but not onto

False; is onto but not one-to-one

Correct answer:

False; is neither one-to-one nor onto

Explanation:

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

T(A) = O , then A = O

Thus, we can prove that is not one-to-one - and not onto - by finding a nonzero column matrix such that T(A)= XA = O .

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

T(A)= XA

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

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

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

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

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

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Example Question #52 : Linear Mapping

is the set of all polynomials of finite degree in .

Define a linear mapping $T:P \rightarrow P$ as follows:

$T(p(x)) =\frac{\mathrm{d} ^2}{\mathrm{d} x^2}p(x)$ .

True or false: T(x) is a one-to-one and onto linear mapping.

Possible Answers:

True

False: is onto but not one-to-one.

False: is neither one-to-one nor onto.

False: is one-to-one but not onto.

Correct answer:

False: is onto but not one-to-one.

Explanation:

The domain and the codomain are both of infinite dimension, so it is possible for be one-to-one, onto, both, or neither.

is one-to-one if and only if

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

Let p(x)= x+ 1 and q(x)= x+ 2

Then

$T(p)= \frac{\mathrm{d} ^2}{\mathrm{d} x^2}p(x)= 0$ and $T(q)= \frac{\mathrm{d} ^2}{\mathrm{d} x^2}q(x)= 0$ .

Since

T(p) = T(q) , but $p \ne q$ , is not one-to-one.

Now let $t(x)= \sum_{n= 0}^{\infty} a_{n}x^{n}$ , where finitely many $a_{n}$ are nonzero.If

$p (x)= \sum_{n= 0}^{\infty} \frac{a_{n}}{(n+2)(n+1)}x^{n+2}$ ,

then

$T(p)= \frac{\mathrm{d} ^2}{\mathrm{d} x^2} \left ( \sum_{n= 0}^{\infty} \frac{a_{n}}{(n+2)(n+1)}x^{n+2} \right )$

$= \sum_{n= 0}^{\infty} a_{n}x^{n} = t(x)$

is therefore onto.

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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 #904 : Linear Algebra

Example Question #905 : Linear Algebra

Example Question #906 : Linear Algebra

Example Question #51 : Linear Mapping

Example Question #52 : Linear Mapping

All Linear Algebra Resources

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