Bijection Transformations: Is T A Bijection?

Let's dive into determining whether the given transformations are bijections. A bijection, as you might already know, is a function that is both injective (one-to-one) and surjective (onto). In simpler terms, it's a perfect pairing between two sets: every element in the first set maps to a unique element in the second set, and every element in the second set has a unique element in the first set mapping to it. We'll analyze each transformation to check for injectivity and surjectivity.

Transformation i: $T${\binom{x}{y}}$=\binom{x+y}{2 x-y}$

We are given the transformation $T${\binom{x}{y}}$=\binom{x+y}{2 x-y}$. To determine if this transformation is a bijection, we must prove it is both injective (one-to-one) and surjective (onto).

Injectivity (One-to-One)

To prove injectivity, we need to show that if $T${\binom{x_1}{y_1}}$ = T${\binom{x_2}{y_2}}$$, then $\binom{x_1}{y_1} = \binom{x_2}{y_2}$. Let's assume that

$T${\binom{x_1}{y_1}}$ = T${\binom{x_2}{y_2}}$$

This implies

$\binom{x_1+y_1}{2x_1-y_1} = \binom{x_2+y_2}{2x_2-y_2}$

So we have the following system of equations:

$x_1 + y_1 = x_2 + y_2$ ...(1)

$2x_1 - y_1 = 2x_2 - y_2$ ...(2)

Adding equations (1) and (2), we get:

$3x_1 = 3x_2$, which implies $x_1 = x_2$.

Substituting $x_1 = x_2$ into equation (1), we get:

$x_1 + y_1 = x_1 + y_2$, which implies $y_1 = y_2$.

Since $x_1 = x_2$ and $y_1 = y_2$, we have $\binom{x_1}{y_1} = \binom{x_2}{y_2}$. Therefore, the transformation T is injective.

Surjectivity (Onto)

To prove surjectivity, we need to show that for any vector $\binom{u}{v}$ in the codomain, there exists a vector $\binom{x}{y}$ in the domain such that $T${\binom{x}{y}}$ = \binom{u}{v}$. In other words, we need to find x and y such that:

$x + y = u$ ...(3)

$2x - y = v$ ...(4)

Adding equations (3) and (4), we get:

$3x = u + v$, which implies $x = \frac{u + v}{3}$.

Substituting $x = \frac{u + v}{3}$ into equation (3), we get:

$\frac{u + v}{3} + y = u$, which implies $y = u - \frac{u + v}{3} = \frac{3u - u - v}{3} = \frac{2u - v}{3}$.

So, for any $\binom{u}{v}$, we can find $\binom{x}{y} = \binom{\frac{u+v}{3}}{\frac{2u-v}{3}}$ such that $T${\binom{x}{y}}$ = \binom{u}{v}$. Therefore, the transformation T is surjective.

Since T is both injective and surjective, it is a bijection. Great job understanding this first transformation! Remember, the key here is to rigorously demonstrate that every input maps to a unique output (injectivity) and that every possible output has a corresponding input (surjectivity). The algebraic manipulation is crucial to finding these relationships. Keep practicing these techniques and you'll master them in no time! We meticulously checked that each vector in the target space has a pre-image, and that distinct vectors in the original space map to distinct vectors in the target space. This is the essence of a bijection! Don't forget, understanding these concepts will be super helpful in various areas of math and computer science where transformations are key.

Transformation ii: $T${\left(\begin{array}{l}a \ b \ c\end{array}\right)}$=\binom{a-b-c}{-a+b}$

Now, let's analyze the second transformation: $T${\left(\begin{array}{l}a \ b \ c\end{array}\right)}$=\binom{a-b-c}{-a+b}$. Again, we need to determine if this transformation is both injective and surjective.

Injectivity (One-to-One)

To prove injectivity, we need to show that if $T${\begin{pmatrix} a_1 \ b_1 \ c_1 \end{pmatrix}}$ = T${\begin{pmatrix} a_2 \\ b_2 \\ c_2 \end{pmatrix}}$$, then $\begin{pmatrix} a_1 \\ b_1 \\ c_1 \end{pmatrix} = \begin{pmatrix} a_2 \\ b_2 \\ c_2 \end{pmatrix}$. Let's assume that

$T${\begin{pmatrix} a_1 \ b_1 \ c_1 \end{pmatrix}}$ = T${\begin{pmatrix} a_2 \\ b_2 \\ c_2 \end{pmatrix}}$$

This implies

$\binom{a_1 - b_1 - c_1}{-a_1 + b_1} = \binom{a_2 - b_2 - c_2}{-a_2 + b_2}$

So we have the following system of equations:

$a_1 - b_1 - c_1 = a_2 - b_2 - c_2$ ...(5)

$-a_1 + b_1 = -a_2 + b_2$ ...(6)

From equation (6), we have $b_1 = a_1 - a_2 + b_2$. Substituting this into equation (5), we get:

$a_1 - (a_1 - a_2 + b_2) - c_1 = a_2 - b_2 - c_2$

$a_1 - a_1 + a_2 - b_2 - c_1 = a_2 - b_2 - c_2$

$a_2 - b_2 - c_1 = a_2 - b_2 - c_2$

$-c_1 = -c_2$, which implies $c_1 = c_2$.

Now, we have $b_1 = a_1 - a_2 + b_2$ and $c_1 = c_2$. Let's consider a counterexample to show that the transformation is not injective. Suppose $a_1 = 1$, $b_1 = 1$, $c_1 = 0$ and $a_2 = 0$, $b_2 = 0$, $c_2 = 0$. Then,

$T${\begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}}$ = \binom{1 - 1 - 0}{-1 + 1} = \binom{0}{0}$

$T${\begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}}$ = \binom{0 - 0 - 0}{-0 + 0} = \binom{0}{0}$

So, $T${\begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}}$ = T${\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}}$$, but $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \neq \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$. Therefore, the transformation T is not injective.

Surjectivity (Onto)

To prove surjectivity, we need to show that for any vector $\binom{u}{v}$ in the codomain, there exists a vector $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ in the domain such that $T${\begin{pmatrix} a \ b \ c \end{pmatrix}}$ = \binom{u}{v}$. In other words, we need to find a, b, and c such that:

$a - b - c = u$ ...(7)

$-a + b = v$ ...(8)

From equation (8), we have $b = a + v$. Substituting this into equation (7), we get:

$a - (a + v) - c = u$

$a - a - v - c = u$

$-v - c = u$, which implies $c = -u - v$.

So, we have $b = a + v$ and $c = -u - v$. We can choose a to be any real number. For example, let $a = 0$. Then $b = v$ and $c = -u - v$. So, for any $\binom{u}{v}$, we can find $\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ v \\ -u-v \end{pmatrix}$ such that $T${\begin{pmatrix} a \ b \ c \end{pmatrix}}$ = \binom{u}{v}$. Therefore, the transformation T is surjective.

Since T is not injective but is surjective, it is not a bijection. That's a wrap on our second transformation! Remember, to be a bijection, a transformation must be both injective and surjective. Even if it satisfies one condition, it's not enough. This example highlights the importance of checking both properties when determining if a transformation is a bijection.

In summary:

i. $T${\binom{x}{y}}$=\binom{x+y}{2 x-y}$ is a bijection.

ii. $T${\left(\begin{array}{l}a \ b \ c\end{array}\right)}$=\binom{a-b-c}{-a+b}$ is not a bijection.

Keep practicing, and you'll become a pro at determining whether transformations are bijections! Remember the key concepts: injectivity (one-to-one) and surjectivity (onto). Analyze each transformation systematically, and you'll ace it every time. Good job, guys! Understanding bijections opens doors to more advanced mathematical concepts, so keep up the great work!