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Determine which of the following transformations are linear transformations

A. The transformation $T_1$ defined by $T_1(x_1,x_2,x_3)=(x_1,0,x_3)$

B. The transformation $T_2$ defined by $T_2(x_1,x_2)=(2x_1−3x_2,x_1+4,5x_2)$.

C. The transformation $T_3$ defined by $T_3(x_1,x_2,x3)=(x_1,x_2,−x_3)$

D. The transformation $T_4$ defined by $T_4(x_1,x_2,x3)=(1,x_2,x_3)$

E. The transformation $T_5$ defined by $T_5(x_1,x_2)=(4x_1−2x_2,3|x_2|)$.

I believe that it could be A and E. How can I determine this? If someone could show me one I could figure out the rest.

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6 Answers 6

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To test whether T is a linear transformation, you need to check that for some vectors $a$ and $b$ and some constant $c$
$$T(a + b) = T(a) +T(b)$$ $$T(ca) = cT(a)$$ $$T(0) = 0$$ So for example,
A. $T(x_1,x_2,x_3)=(x_1,0,x_3)$ $$T(x_1+y_1,x_2+y_2,x_3+y_3)=(x_1+y_1,0(x_2+y_2),x_3+y_3)=T(x_1,0,x_3)+T(y_1,0,y_2)$$$$T(cx_1,cx_2,cx_3)=T(cx_1,(c)0,cx_3)=cT(x_1,0,x_3)$$ $$T(0,0,0)=0$$ B. $T(x_1,x_2)=(2x_1−3x_2,x_1+4,5x_2)$ $$T(x_1+y_1,x_2+y_2)=(2(x_1+y_1)−3(x_2+y_2),(x_1+y_1)+4,5(x_2+y_2))=(2x_1+2y_1−3x_2-3y_2,x_1+y_1+4,5x_2+5y_2)$$$$T(x_1,x_2)+T(y_1,y_2)=(2x_1−3x_2,x_1+4,5x_2)+(2y_1−3y_2,y_1+4,5y_2)=(2x_1−3x_2+2y_1-3y_2,x_1+y_1+8,5x_2+5y_2)\not=T(x_1+y_1,x_2+y_2)$$ So B is not a linear transformation.

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  • $\begingroup$ What about $T(a,b,c)$ ? $\endgroup$
    – Yusha
    Jun 20, 2016 at 20:29
  • $\begingroup$ In the example above $a, b$ are vectors and $c$ is a scalar. $\endgroup$
    – Doug M
    Jun 20, 2016 at 20:31
  • $\begingroup$ @Yusha It is not necessarily following the format in your problem, it is asking you to plug in two vectors and see if they work according to the transformation given, so it doesn't matter how many terms are there in your given T $\endgroup$
    – Bridget
    Jun 20, 2016 at 20:35
  • $\begingroup$ Can you do B also? I just want to see one more $\endgroup$
    – Yusha
    Jun 20, 2016 at 20:37
  • $\begingroup$ @Yusha just added B $\endgroup$
    – Bridget
    Jun 20, 2016 at 21:03
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For $T$ to be a linear transformation, two criteria need to be satisfied:

  1. $T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y})$
  2. $T(a\mathbf{x}) = aT(\mathbf{x})$ for $a$ a scalar/constant.

As an example, suppose $\mathbf{x} = (x_1, x_2, x_3)$ and $\mathbf{y} = (y_1, y_2, y_3)$. Let's start with A.

Then, $T(\mathbf{x}) = (x_1, 0, x_3)$ and $T(\mathbf{y}) = (y_1, 0, y_3)$. Now $$\mathbf{x}+\mathbf{y} = (x_1 + y_1, x_2 + y_2, x_3 + y_3)\text{.}$$ It follows that $T(\mathbf{x}+\mathbf{y}) = (x_1 + y_1, 0, x_3+y_3)$.

Is $T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x})+T(\mathbf{y})$?

Now, for a constant $a$, $a\mathbf{x} = (ax_1, ax_2, ax_3)$.

We have $T(a\mathbf{x}) = (ax_1, 0, ax_3)$.

Furthermore, we know $T(\mathbf{x}) = (x_1, 0, x_3)$, so $aT(\mathbf{x}) = (ax_1, 0, ax_3)$.

Is $T(a\mathbf{x}) = aT(\mathbf{x})$?

Repeat this for all of the other $T$.

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A linear transformation has this defintion.

$T(\mathbf x+\mathbf y) = T(\mathbf x) + T(\mathbf y)\\ T(c\mathbf x) = cT(\mathbf x)$

Show that this is (or isn't) true for each of the above.

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    $\begingroup$ that's an additive transformation -- it doesn't necessarily scale appropriately $\endgroup$
    – Chill2Macht
    Jun 20, 2016 at 20:26
  • $\begingroup$ @William It is adequate in $\mathbb R^n,$ as $T(\mathbf x +\mathbf x) = 2 T(\mathbf x).$ But, yes, $T(c\mathbf x) = cT(\mathbf x)$ should be included as long as I am quoting definitions. $\endgroup$
    – Doug M
    Jun 20, 2016 at 20:36
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    $\begingroup$ True, but that will only work if the underlying field is complete, and even then it is a tedious pain to show how it is sufficient in that special case. It is technically correct, but I don't think it is pedagogically useful for someone learning abstract vector spaces for the first time. $\endgroup$
    – Chill2Macht
    Jun 20, 2016 at 20:39
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    $\begingroup$ @William, it also needs the fraction field of the image of $Z$ to be dense, I would think? Correct me if I'm wrong, but in that case, a complete non-discrete field of characteristic $p$ should be a problem. See wikiwand.com/en/Local_field for existence of such fields. Please let me know if I'm wrong, so I can think about it more. $\endgroup$
    – jgon
    Jun 21, 2016 at 3:14
  • $\begingroup$ @jgon You're right of course; my bad $\endgroup$
    – Chill2Macht
    Jun 21, 2016 at 3:47
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$A$. $$T(a\vec{x})=T(ax_1, ax_2, ax_3) = (ax_1, 0, ax_3) = a(x_1, 0, x_3) = aT(\vec{x})$$

$$T(\vec{x} +\vec{y})=T(x_1+y_1, x_2+y_2, x_3+y_3)\\=(x_1+y_1,0,x_3+y_3)=(x_1,0,x_3)+(y_1,0,y_3)=T(\vec{x})+T(\vec{y})$$

Therefore the first transformation is linear as you correctly guessed. Just repeat the same procedure for B-E and see if it works or not.

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A linear transformation is defined by a set of homogeneous linear (i.e. all terms have degree $1$) polynomials, namely of the form $$T(x_1,x_2,x_3)=ax_1+bx_2+cx_3\quad(a,b,c\in\mathbf R).$$ A and C alone satisfy this criterion.

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  • $\begingroup$ That's a linear transformation from $\mathbb{R}^3 \to \mathbb{R}$; not a linear endomorphism of $\mathbb{R}^3$ $\endgroup$
    – Chill2Macht
    Jun 20, 2016 at 20:30
  • $\begingroup$ @William: I rephrased my answer in a more appropriate manner. $\endgroup$
    – Bernard
    Jun 20, 2016 at 20:38
  • $\begingroup$ @Bernard I still agree with William, unless $a$, $b$, and $c$ are vectors (which is not what you specified), and then it isn't a homogeneous linear polynomial. You could say it is a polynomial map with each coordinate function being a homogeneous linear polynomial. $\endgroup$
    – jgon
    Jun 21, 2016 at 3:09
  • $\begingroup$ @jgon: That was implicit when I added ‘defined by a set of (one polynomial per coordinate of the codomain). $\endgroup$
    – Bernard
    Jun 21, 2016 at 8:44
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Is T in A a linear transformation?

  1. Check linearity for addition.

Suppose $T:V \rightarrow W$ .Where $V$ and $W$ are vector spaces over $F$. Let $x_1,x_2,x_3 \in F$ and also let $x_4,x_5,x_6 \in F$. So that $(x_1,x_2,x_3) \in V$ and $(x_4,x_5,x_6) \in V$. Now need to check that $T((x_1,x_2,x_3)) + T((x_4,x_5,x_6)) = T((x_1+x_4, x_2+x_5,x_3+x_6))$.

We have LHS $ = (x_1,0,x_3) + (x_4,0,x_6) = (x_1+x_4,0,x_3+x_6) = $RHS. Hence this holds by definition of vector addition.

  1. Check linearity for scalar multiplication:

Let be as above, and suppose $V$ is a vector-space over a field $F$. Then let $a\in F$. Want to prove that:

$aT(x_1,x_2,x_3)=T(ax_1,ax_2,ax_3)$. Hence we have:

LHS $ = a(x_1,0,x_3) = (ax_1,0,ax_3) = $RHS. And this holds by vector scalar multiplication and by property of zero in $\mathbb{R}$.

Hence this is a linear transformation by definition. In general you need to show that these two properties hold.

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  • $\begingroup$ If $x_1,x_2,x_3$ are vectors in $V$ and $T:V\rightarrow W$, what is $(x_1,x_2,x_3)$ and how do you define $T$ on something which is clearly not in the domain? $\endgroup$
    – sqtrat
    Jun 20, 2016 at 20:42
  • $\begingroup$ ahaha you're right sorry $\endgroup$
    – Euler_Salter
    Jun 20, 2016 at 20:43
  • $\begingroup$ @sqtrat , edited now! $\endgroup$
    – Euler_Salter
    Jun 20, 2016 at 20:45
  • $\begingroup$ It seems that you assume that $V$ and $W$ are $3$ dimensional with standard basis vectors, otherwise why should $(x_1,x_2,x_3)$ be an element of $V$ in the first place. Either be more general or be more specific. $\endgroup$
    – sqtrat
    Jun 20, 2016 at 20:51
  • $\begingroup$ @sqtrat could they be more than 3 dimensional? $\endgroup$
    – Euler_Salter
    Jun 20, 2016 at 20:53

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