Properties of linear transformation

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Intros
Lessons
  1. Properties of Linear Transformation Overview:
  2. The 3 properties of Linear Transformation
    T(u+v)=T(u)+T(v)T(u+v)=T(u)+T(v)
    T(cu)=cT(u)T(cu)=cT(u)
    T(0)=0T(0)=0
  3. How to see if a transformation is linear
    • Show that: T(cu+dv)=cT(u)+dT(v)T(cu+dv)=cT(u)+dT(v)
    • General formula: T(c1v1+c2v2++cnvn)=c1T(v1)+c2T(v2)++cpT(vp)T(c_1 v_1+c_2 v_2+\cdots+c_n v_n )=c_1 T(v_1 )+c_2 T(v_2 )+\cdots+c_p T(v_p)
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Examples
Lessons
  1. Understanding and Using the Properties
    Show that the transformation TT defined by understand properties of linear transformation is not linear.
    1. Show that the transformation TT defined by prove linear transformation is not linear.
      1. Proving Questions using the Properties
        An affine transformation T:RnT: \Bbb{R}^n Rm \Bbb{R}^m has the form T(x)=Ax+bT(x)=Ax+b, where AA is an m×nm \times n matrix and bb is a vector in Rn\Bbb{R}^n. Show that the transformation TT is not a linear transformation when b0b \neq 0.
        1. Define T:RnT: \Bbb{R}^n Rm \Bbb{R}^m to be a linear transformation, and let the set of vectors {v1,v2,v3v_1,v_2,v_3 } be linearly dependent. Show that the set of vectors {T(v1),T(v2),T(v3)T(v_1),T(v_2),T(v_3)} are also linearly dependent.
          1. Define T:RnT: \Bbb{R}^n Rm \Bbb{R}^m to be a linear transformation and the set of vectors v1v_1,...,vpv_p are in Rn\Bbb{R}^n. In addition, let T(vi)=0T(v_i )=0 for i=1,2,i=1,2,,p,p. If xx is any vector in Rn\Bbb{R}^n, then show that T(x)=0T(x)=0. In other words, show that TT is the zero transformation.
            Topic Notes
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            Properties of linear transformation


            What is a linear transformation

            A transformation is a technique in which a vector gets converted into another by keeping a unique element from each of the original vector and assigning it into the resulting vector. This process basically maps one vector space into another. There may be different kinds of transformations in mathematics but the case of a linear transformation in linear algebra is that which preserves linear combinations from the original expression to the resulting one, in simple words, it preserves the way addition and scalar multiplication operate from the original vector to the transformed vector. A linear transformation example can also be called linear mapping since we are keeping the original elements from the original vector and just creating an image of it.

            Recall the matrix equation Ax = b, normally, we say that the product of AA and xx gives bb. Now we are going to say that AA is a linear transformation matrix that transforms a vector xx into a vector bb (we now call b an image of a linear transformation of xx).

            Properties of linear transformation

            In a sense AxA{x} is a function where if we plug in a vector, then it spits out another vector. If we call this function T(x)T(x), then T(x)=AxT(x)=Ax, where TT is the transformation. Note that T(x)T(x) is an image of xx since T(x)=bT(x)=b;

            How to do a linear transformation

            From the linear transformation definition we have seen above, we can plainly say that to perform a linear transformation or to find the image of a vector xx, is just a fancy way to say "compute T(x)T(x)".

            So to compute a linear transformation is to find the image of a vector, and it can be any vector, therefore:

            T(x)T(x) = Image of xx under the transformation TT

            T(v)T(v) = Image of vv under the transformation TT

            And so, if we define TT: R2R^{2} \, R2 \, R^{2} by T(x)=AxT(x)=Ax. Find the image of vv under TT if:

            Properties of linear transformation
            Equation 1: Matrix and vector to perform transformation

            So our goal is to find T(v)=AvT(v)=Av
            For that, remember our matrix multiplication guide:

            Properties of linear transformation
            Equation 2: Matrix multiplication


            And so, we perform the transformation:

            Properties of linear transformation
            Equation 3: Linear transformation of v


            Therefore the image of vv is:

            Properties of linear transformation
            Equation 4: Image of v

            But what if the question gives the image and asks you to find the original vector? Let us use the same vectors as above and think how can we do the reverse engineering to obtain vv, we start by working out the matrix equation below:

            Properties of linear transformation
            Equation 5: Matrix equation for the linear transformation of v

            We use what we learned on the lesson about representing a linear system as a matrix, and write down an augmented matrix from our matrix equation above. Then in order to solve for vector v we just have to use row reduction and echelon forms!

            Properties of linear transformation
            Equation 6: Solving for vector v

            And as you can see, we just got the linear transformation proof by reversing the process and obtaining the same vector v we started with:

            Properties of linear transformation
            Equation 7: Vector v

            Properties of Linear transformations

            There are many linear transformation applications, but before start working on problems where we can use them, it is time to look into their properties. On our last lesson we talked about the image and the range of linear transformation T(x)T(x), now let us see how the operations of addition/subtraction and multiplication apply to them.

            Recall from last chapter the 2 properties of AxAx:
            1. A(u+v)=Au+AvA(u+v) = Au + Av
            2. A(cu)=c(Au)A(cu) = c(Au)

            where uuu and vvv are vectors in RnR^{n} and cc is a scalar.

            Now the properties of linear transformations are very similar. Linear transformation preserves the operations of vector addition/subtraction and scalar multiplication. In other words, If TT is linear, then:
            1. T(u+v)=T(u)+T(v)T(u+v) = T(u) + T(v)
            2. T(cu)=cT(u)T(cu) = cT(u)
            3. T(0)=0T (\rightharpoonup_{0}) = \rightharpoonup_{0}

            We can even combine property 1 and 2 to show that:

            T(cu+dv)=cT(u)+dT(v)T(cu \,+\, dv) = cT(u) \,+ \, dT(v)
            Equation 8: Combination of the properties for a linear transformation

            where uu, vv are vectors and cc, dd are scalars. Note that if this equation holds, then it must be linear.
            What if you have more than 2 vectors and 2 scalars? What if you have p vectors and p scalars? Then we can generalize this equation and say that:

            T(c1v1+c2v2+...+cpvp)=c1T(v1)+c2T(v2)+...+cpT(vp)+T(c_{1}v_{1} \,+\, c_{2}v_{2} \, + \, . . . \, + \, c_{p}v_{p}) = c_{1}T(v_{1}) \,+ \, c_{2}T(v_{2}) \,+ \, . . . \, + \, c_{p}T(v_{p}) \,+ \,
            Equation 9: Linear transformation of p vectors

            Notice that transformations of linear functions are distributive. And again if this equation holds, then it must be linear.

            Example 1

            Show that the transformation TT defined below is not linear:

            Properties of linear transformation
            Equation 10: Linear transformation of vector x

            For this problem, we have to work with the matrix representation of linear transformation T(x)T(x) to prove if the third property is held:

            Properties of linear transformation
            Equation 11: Applying the third property of linear transformations

            As you can see, the linear transformation does not result in a zero vector, and therefore, it cannot be linear.

            Example 2

            Show that the transformation T defined below is not linear:

            Properties of linear transformation
            Equation 12: Linear transformation

            This particular problem gets a little bit more complicated to prove since there is an absolute value inside the matrix of linear transformation TT. Thus, we will use two properties instead of just one, we start with the first property to work through, and then the third property is used in combination while working on the left hand side of the first property to finish this problem:
            We start by selecting our own vectors uu and vv (you can choose any vectors as long as the conditions are met).

            Properties of linear transformation
            Equation 13: Vectors u and v

            And then using property one, we compute the transformations:

            Properties of linear transformation
            Equation 14: Applying the first property of linear transformations

            Notice in the left hand side we have used the third property too, to know the transformation of a zero vector is just the zero vector itself. Thus our result is:

            Properties of linear transformation
            Equation 15: First property of linear transformations not met

            And since the first property is not met, the transformation is not linear.

            Example 3


            An affine transformation T:RnT: R^{n} \, Rm \, R^{m} \, has the form T(x)=Ax+bT(x)=Ax+b, where AA is an m×nm \times n matrix of a linear transformation and bb is a vector in RnR^{n}. Show that the transformation TT is not a linear transformation when bb ≠ 0.

            For this problem we need to check if any of the three properties of a linear transformation do not hold. Starting by the first property, we create the next table in which you can see how the left hand side and the right hand side of property one have been applied to the affine transformation above and see if they are equal when bb ≠ 0.

            T(u+v)T(u+v)

            T(u)+T(v)T(u)+T(v)

            T(u+v)=A(u+v)+T(u+v)=A(u+v)\,+
            b=Au+Av+bb=Au+Av+b

            T(u)+T(v)=(Au+b)+T(u)+T(v)=(Au+b)\,+
            (Av+b)=Au+Av+2b(Av+b)=Au+Av+2b

            Equation 16: Applying the first property of linear transformations

            So as you can see, when bb ≠ 0. the first property of a linear transformation is not met, and that is it, it means this transformation T is not linear.

            Example 4

            Define T:RnT: R^{n} \, Rm \, R^{m} to be a linear transformation, and let the set of vectors {v1,v2,v3v_{1},v_{2},v_{3}} be linearly dependent. Show that the set of vectors {T(v1),T(v2),T(v3)T(v_{1}),T(v_{2}),T(v_{3})} are also linearly dependent.

            This question may sound a little complicated, so let us bring back a quick review from our lesson on linear independence where we learnt that:

            A set of vectors {v1,v2,v3,...,vpv_{1},v_{2},v_{3}, ... , v_{p}} is said to be linearly independent if:

            v1c1+v2c2+v3c3+...+vpcp=0v_{1}c_{1} + v_{2}c_{2} + v_{3}c_{3} + ... + v_{p}c_{p} = 0

            iff:c1+c2+c3+...+cp=0iff: {c_{1} + c_{2} + c_{3} + ... + c_{p}} = 0
            Equation 17: Condition for linear independence

            So, linear independence produces a trivial solution, therefore in this case, since we are looking for a linearly dependant set of vectors, this means that we cannot have all zero solutions.

            Thus, we are looking for c1T(v1)+c2T(v2)+c3T(v3)=0c_{1}T(v_{1}) + c_{2}T(v_{2}) + c_{3}T(v_{3}) = 0 where c1,c2,c3,...,cp{c_{1}, c_{2}, c_{3}, ... , c_{p}} cannot all be zeros.

            In order to show that the set of vectors T(v1),T(v2),T(v3){T(v_{1}),T(v_{2}),T(v_{3})} are also linearly dependent, we will be using the three properties of linear transformations!
            1. T(u+v)=T(u)+T(v)T(u+v) = T(u) + T(v)
            2. T(cu)=cT(u)T(cu) = cT(u)
            3. T(0)=0T (\rightharpoonup_{0}) = \rightharpoonup_{0}

            Therefore:
            v1c1+v2c2+v3c3=0v_{1}c_{1} + v_{2}c_{2} + v_{3}c_{3} = 0

            We replace this into the third property to obtain:

            T(0)=T(v1c1+v2c2+v3c3)=0T(0) = T(v_{1}c_{1} + v_{2}c_{2} + v_{3}c_{3}) = 0

            Now using the first property:

            T(v1c1+v2c2+v3c3)=T(v1c1)+T(v2c2)+T(v3c3)=0T(v_{1}c_{1} + v_{2}c_{2} + v_{3}c_{3}) = T(v_{1}c_{1}) + T(v_{2}c_{2}) + T(v_{3}c_{3}) = 0

            And finally using the second property this becomes:

            T(v1c1)+T(v2c2)+T(v3c3)=c1T(v1)+c2T(v2)+c3T(v3)=0T(v_{1}c_{1}) + T(v_{2}c_{2}) + T(v_{3}c_{3}) = c_{1}T(v_{1}) + c_{2}T(v_{2}) + c_{3}T(v_{3}) = 0
            Equation 18: Showing the transformation vectors are linearly dependent

            So at this point we have arrived to exactly the formula we were looking for and now, we just have to show that c1,c2,c3,...,cp{c_{1}, c_{2}, c_{3}, ... , c_{p}} are not all zeros. The trick here is that since we already know from the beginning that the vectors v1,v2,v3{v_{1},v_{2},v_{3}} are linearly dependent, by combining them as in the very first expression in equation 18 with c1,c2,c3,...,cp {c_{1}, c_{2}, c_{3}, ... , c_{p}} this makes them not all zero!, and so, we have made sure this particular characteristic applies to the whole expression found at the end of equation 18.

            Hence: c1T(v1)+c2T(v2)+c3T(v3)=0 c_{1}T(v_{1}) + c_{2}T(v_{2}) + c_{3}T(v_{3}) = 0 is linearly dependant.

            Example 5

            Define T:RnT: R^{n} \, Rm \, R^{m} to be a linear transformation and the set of vectors v1,...,vpv_{1} ,..., v_{p} are in RnR^{n}. In addition, let T(vi)=0T(v_{i}) = 0 for i=1,2,...,pi = 1,2,..., p. If xx is any vector in RnR^{n}, then show that T(x)=0T(x) = 0 . In other words, show that TT is the zero transformation. If we have that:

            x=v1c1+v2c2+...+vpcpx = v_{1}c_{1} + v_{2}c_{2} + ...+ v_{p}c_{p}
            Equation 19: Defining vector x

            Computing the transformation of xx:

            T(x)=T(v1c1+v2c2+...+vpcp)=T(v1c1)+T(v2c2)+...+T(vpcp)T(x) = T(v_{1}c_{1} + v_{2}c_{2} +...+ v_{p}c_{p}) = T(v_{1}c_{1}) + T(v_{2}c_{2}) +...+ T(v_{p}c_{p})

            T(x)=c1T(v1)+c2T(v2)+...+cpT(vp)T(x) = c_{1}T(v_{1}) + c_{2}T(v_{2}) +...+ c_{p}T(v_{p})
            Equation 20: Linear transformation of x


            Using the condition provided in the problem T(vi)=0T(v_{i}) = 0 , this means that:

            T(v1)=0T(v_{1}) = 0

            T(v2)=0T(v_{2}) = 0

            \downarrow

            T(vp)=0T(v_{p}) = 0
            Equation 21: Initial condition

            Then:

            T(x)=c1T(v1)+c2T(v2)+...+cpT(vp)=c1(0)+c2(0)+...+cp(0)=0T(x)=c_{1}T(v_{1})+c_{2}T(v_{2}) +... + c_{p}T(v_{p}) = c_{1}(0)+c_{2}(0) +...+ c_{p}(0) = 0
            Equation 22: Final solution


            And so, we just showed that T(x)=0T(x) = 0 !

            ***

            To continue your practice on linear transformations there are a few links we would like to recommend you such as this in depth calculus III lesson, of this article on linear transformations.

            This finalizes our lesson for today, we hope you enjoyed it, see you in the next one!
            Recall from last chapter the 2 properties of AxAx:
            1. A(u+v)=Au+AvA(u+v)=Au+Av
            2. A(cu)=c(Au)A(cu)=c(Au)

            where uu and vv are vectors in Rn\Bbb{R}^n and cc is a scalar.

            Now the properties of linear transformation are very similar. Linear transformation preserves the operations of vector addition/subtraction and scalar multiplication. In other words, If T is linear, then:
            1. T(u+v)=T(u)+T(v)T(u+v)=T(u)+T(v)
            2. T(cu)=cT(u)T(cu)=cT(u)
            3. T(0)=0T(\vec{0})=\vec{0}

            We can even combine property 1 and 2 to show that:

            T(cu+dv)=cT(u)+dT(v)T(cu+dv)=cT(u)+dT(v)

            where uu, vv are vectors and cc, dd are scalars. Note that if this equation holds, then it must be linear.

            If you have more than 2 vectors and 2 scalars? What if you have p vectors and p scalars? Then we can generalize this equation and say that:

            T(c1v1+c2v2++cpvp)=c1T(v1)+c2T(v2)++cpT(vp) T(c_1 v_1+c_2 v_2+\cdots+c_p v_p )=c_1 T(v_1 )+c_2 T(v_2 )+\cdots+c_p T(v_p)

            Again if this equation holds, then it must be linear.