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Linear Algebra : Linear Algebra

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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:

False;is one-to-one but not onto

False;is onto but not one-to-one

False;is neither one-to-one nor onto

True

Correct answer:

False;is neither one-to-one nor onto

Explanation:

The domain and the codomain ofare identical, sois one to one if and only if it is onto.

A necessary and sufficient condition forto be one-to-one is that the kernel ofbe $\left \{ 0 \right \}$ . 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 thatis not one-to-one - and not onto - by finding a nonzero column matrixsuch 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, sois 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:

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

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

True

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 forbe 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) ,但 $p \ne q$ ,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 : Linear Algebra

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Example Question #51 :Linear Mapping

Example Question #52 :Linear Mapping

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