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Linear Algebra With Applications
Found in: Page 184
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Find out which of the transformations in Exercises 1 through 50 are linear. For those that are linear, determine whether they are isomorphism, T(x+iy)=x2+y2 from to .

The transformation T(x+iy)=x2+y2 is a linear transformation and is an isomorphism.

See the step by step solution

Step by Step Solution

Step 1: Definition of Linear Transformation

Consider two linear spaces V and W . A transformation T is said to be a linear transformation if it satisfies the properties,

T(f+g)=T(f)+T(g) T(kv)=kT(v)

For all elements f,g of v and is scalar.

An invertible linear transformation is called an isomorphism.

Step 2: Check whether the given transformation is a linear or not.

Consider the transformation T(x+iy)=x2+y2, from to .

Check whether the transformation satisfies the below two conditions or not.

1.T(A+B)=T(A)+T(B) 2.T(kA)=kT(A)

Verify the first condition.

Let A=x1+iy1 and B=x2+iy2 be arbitrary complex numbers from . Then,

T(A+B)=T(x1+iy1+x2+iy2) =Tx1+x2+iy1+y2 =ix1+x2+iy1+y2 =ix1+x2+iy1+iy2 =ix1+iy1+x2+iy2 =ix1+iy1+ix2+iy2T(A+B)=T(A)+T(B)

It is clear that, the first condition T(A+B)=T(A)+T(B) is satisfied.

Verify the second condition.

Let k be an arbitrary scalar, and A as follows.

T(kA)=Tkx1+iy1 =ikx1+iy1 =kix1+iy1 =kTx1+iy1TkA=kT(A)

It is clear that, the second condition TkA=kT(A) is also satisfied.

Thus,T is a linear transformation.

Step 3: Properties of isomorphism

A linear transformation T:VW is said to be an isomorphism if and only if ker(T)={0} and Im(T)=W.

Now, check whether ker (T) ={0}.

According to the definition of the kernel of a transformation,

localid="1659422383534" ker(T)={A ,T(A)=0}.

Consider a complex number A as A = x + iy


T(A)=0 T(x+iy)=0 i(x+iy)=0 ix(x+iy)=0 ix+i2y=0+0i ix-y=0+0i -y+ix=0+0i

Comparing both sides, it can be concluded that x = 0 , y = 0

Clearly, ker(T)={0}.

Now, check whether Im (T) =.

According to the definition of the image of a transformation,

localid="1659416713747" Im(T)=T(A):A

Let A=x+iy is in .


T(A)=T(x+iy) =i(x+iy) =ix+i2y =ix-y =-y+ix

, say (B = - Y + ix).

T (A) = B

It is clear that, T(A)=-y+ix.

This means for any localid="1659422430378" B there exists a localid="1659422483414" Asuch that T(A)=B.

So, Im (T) =

Therefore, the transformation T is an isomorphism.

Thus, the transformation T is a linear transformation and T is an isomorphism.

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