# 17.1 Determine if T is a linear transformation

• MHB
• karush
In summary, we have discussed a linear transformation $T: \Bbb{R}^2 \to \Bbb{R}^2$ defined by $T \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 2x+y \\ x-4y \end{bmatrix}$. We have verified that $T$ is indeed a linear transformation by showing that $T(\vec{x}+\vec{y})=T(\vec{x})+T(\vec{y})$ and $T(c\vec{x})=cT(\vec{x})$ for all vectors $\vec{x}$ and $\vec{y}$ and scalar $c$.
karush
Gold Member
MHB
nmh{2000}
17.1 Let $T: \Bbb{R}^2 \to \Bbb{R}^2$ be defined by
$$T \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 2x+y\\x-4y \end{bmatrix}$$
Determine if $T$ is a linear transformation. So if
$$T(\vec{x}+\vec{y})=T(\vec{x})+T(\vec{y})$$
Let $\vec{x}$ and $\vec{y}$ be vectors in $\Bbb{R}^2$ then we can write them as
$$\vec{x} =\begin{bmatrix} x_1\\x_2 \end{bmatrix} , \vec{y} =\begin{bmatrix} y_1\\y_2 \end{bmatrix}$$
By definition, we have that
$$T(\vec{x}+\vec{y}) =\begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix} =\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$$

OK just seeing if this is developing as it should
hopefully the next few steps will be an addition property
and this is a linear transformation

Last edited:
karush said:
17.1 Let $T: \Bbb{R}^2 \to \Bbb{R}^2$ be defined by
$$T \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 2x+y\\x-4y \end{bmatrix}$$
Determine if $T$ is a linear transformation. So if
$$T(\vec{x}+\vec{y})=T(\vec{x})+T(\vec{y})$$
Let $\vec{x}$ and $\vec{y}$ be vectors in $\Bbb{R}^2$ then we can write them as
$$\vec{x} =\begin{bmatrix} x_1\\x_2 \end{bmatrix} , \vec{y} =\begin{bmatrix} y_1\\y_2 \end{bmatrix}$$
By definition, we have that
$$T(\vec{x}+\vec{y}) = \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix} =\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$$

OK just seeing if this is developing as it should
hopefully the next few steps will be an addition property
and this is a linear transformation
$$\displaystyle T(\vec{x}+\vec{y}) = T \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix} =\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$$

Good so far. Now what is T(x) + T(y)? And how do you verify T(cx) = cT(x)?

-Dan

ok sorry I got to finish this later
but where is the typo...actually I would say that could be determined just by observation... or not?

karush said:
ok sorry I got to finish this later
but where is the typo...actually I would say that could be determined just by observation... or not?
It's a little bit of nothing. You left the T off in front of the $$\displaystyle \left [ \begin{matrix} x_1 + y_1 \\ x_2 + y_2 \end{matrix} \right ]$$ in the second step of the last equation. (I know there's a way to change the color in LaTeX to highlite it but I'm too lazy to look it up right now.)

-Dan

$\displaystyle T(\vec{x}+\vec{y}) = T \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix} =T\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$
So then
$T(\vec{x})+T(\vec{y}) = T\begin{bmatrix} x_1\\x_2 \end{bmatrix} +T\begin{bmatrix} y_1\\y_2 \end{bmatrix} =\begin{bmatrix} 2(x_1+x_2)\\x_1+x_2\end{bmatrix}+\begin{bmatrix} y_1+y_2\\-4(y_1+y_2 )\end{bmatrix} =\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$
and
$T(c\vec{x})=\begin{bmatrix}c x_1\\cx_2 \end{bmatrix} = \begin{bmatrix} cx_1+cy_1 \\ cx_2+cy_2 \end{bmatrix}$
and
$cT(\vec{x})=c\begin{bmatrix} x_1\\x_2 \end{bmatrix} = c\begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix} = \begin{bmatrix} cx_1+cy_1 \\ cx_2+cy_2 \end{bmatrix}$ok typos for sure

karush said:
$\displaystyle T(\vec{x}+\vec{y}) = T \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \end{bmatrix} =T\begin{bmatrix} 2(x_1+y_1)+x_2+y_2\\ x_1+y_1-4(x_2+y_2) \end{bmatrix}$
Only one typo: Top line of equations, third term doesn't need the T. You've already taken the transform.

The second line of equations needs some work. You have the correct answer, but the intermediate step is wrong. Is this also a typo?

It should be
$$\displaystyle T \left [ \begin{matrix} x_1 \\ x_2 \end{matrix} \right ] + T \left [ \begin{matrix} y_1 \\ y_2 \end{matrix} \right ] = \left [ \begin{matrix} 2x_1 + x_2 \\ x_1 - 4x_2 \end{matrix} \right ] + \left [ \begin{matrix} 2y_1 + y_2 \\ y_1 - 4y_2 \end{matrix} \right ] = \left [ \begin{matrix} 2(x_1 + y_1) + (x_2 + y_2) \\ (x_1 + y_1) - 4(x_2 + y_2) \end{matrix} \right ]$$

-Dan

ok really appreciate the help.

I still need more practice on these...

sorry for the wait on reply but I am limited to when the computer labs are open.

Mahalo...

## 1. What is a linear transformation?

A linear transformation is a mathematical function that maps one vector space to another while preserving the basic structure of the original space. In simpler terms, it is a transformation that maintains the properties of linearity, such as preserving parallel lines and the origin.

## 2. How can I determine if T is a linear transformation?

To determine if T is a linear transformation, you need to check if it satisfies the two properties of linearity: additivity and homogeneity. Additivity means that T(u + v) = T(u) + T(v) for any vectors u and v, and homogeneity means that T(ku) = kT(u) for any scalar k and vector u. If T satisfies these properties, it is a linear transformation.

## 3. What is the difference between a linear transformation and a non-linear transformation?

A linear transformation preserves the properties of linearity, while a non-linear transformation does not. This means that a linear transformation will map parallel lines to parallel lines and the origin to the origin, while a non-linear transformation may distort these properties.

## 4. Can a linear transformation have a non-linear inverse?

No, a linear transformation must have a linear inverse. This is because a linear transformation preserves the properties of linearity, so its inverse must also preserve these properties. If a transformation is non-linear, it cannot have a linear inverse.

## 5. How can I use determinants to determine if T is a linear transformation?

If the determinant of the transformation matrix is non-zero, then the transformation is a linear transformation. This is because the determinant represents the scaling factor of the transformation, and a non-zero scaling factor indicates that the transformation preserves the properties of linearity.

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