5.8: The Matrix of a Linear Transformation II
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- May 12, 2023
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( \newcommand{\kernel}{\mathrm{null}\,}\)
Outcomes
- Find the matrix of a linear transformation with respect to general bases.
We begin this section with an important lemma.
Lemma 5.8.1:Mapping of a Basis
Let T:Rn↦Rn be an isomorphism. Then T maps any basis of Rn to another basis for Rn.
Conversely, if T:Rn↦Rn is a linear transformation which maps a basis of Rn to another basis of Rn, then it is an isomorphism.
- Proof
-
First, suppose T:Rn↦Rn is a linear transformation which is one to one and onto. Let {→v1,⋯,→vn} be a basis for Rn. We wish to show that {T(→v1),⋯,T(→vn)} is also a basis for Rn.
First consider why it is linearly independent. Suppose ∑nk=1akT(→vk)=→0. Then by linearity we have T(∑nk=1ak→vk)=→0 and since T is one to one, it follows that ∑nk=1ak→vk=→0. This requires that each ak=0 because {→v1,⋯,→vn} is independent, and it follows that {T(→v1),⋯,T(→vn)} is linearly independent.
Next take →w∈Rn. Since T is onto, there exists →v∈Rn such that T(→v)=→w. Since {→v1,⋯,→vn} is a basis, in particular it is a spanning set and there are scalars bk such that T(∑nk=1bk→vk)=T(→v)=→w. Therefore →w=∑nk=1bkT(→vk) which is in the span{T(→v1),⋯,T(→vn)}. Therefore, {T(→v1),⋯,T(→vn)} is a basis as claimed.
Suppose now that T:Rn↦Rn is a linear transformation such that T(→vi)=→wi where {→v1,⋯,→vn} and {→w1,⋯,→wn} are two bases for Rn.
To show that T is one to one, let T(∑nk=1ck→vk)=→0. Then ∑nk=1ckT(→vk)=∑nk=1ck→wk=→0. It follows that each ck=0 because it is given that {→w1,⋯,→wn} is linearly independent. Hence T(∑nk=1ck→vk)=→0 implies that ∑nk=1ck→vk=→0 and so T is one to one.
To show that T is onto, let →w be an arbitrary vector in Rn. This vector can be written as →w=∑nk=1dk→wk=∑nk=1dkT(→vk)=T(∑nk=1dk→vk). Therefore, T is also onto.
Consider now an important definition.
Definition 5.8.1: Coordinate Vector
Let B={→v1,→v2,⋯,→vn} be a basis for Rn and let →x be an arbitrary vector in Rn. Then →x is uniquely represented as →x=a1→v1+a2→v2+⋯+an→vn for scalars a1,⋯,an.
The coordinate vector of →x with respect to the basis B, written CB(→x) or [→x]B, is given by CB(→x)=CB(a1→v1+a2→v2+⋯+an→vn)=[a1a2⋮an]
Below is a video on change of basis between the standard basis and a nonstandard basis.
Below is a video on change of basis between two nonstandard bases.
Consider the following example.
Example 5.8.1:Coordinate Vector
Let B={[10],[−11]} be a basis of R2 and let →x=[3−1] be a vector in R2. Find CB(→x).
Solution
First, note the order of the basis is important so label the vectors in the basis B as B={[10],[−11]}={→v1,→v2} Now we need to find a1,a2 such that →x=a1→v1+a2→v2, that is: [3−1]=a1[10]+a2[−11] Solving this system gives a1=2,a2=−1. Therefore the coordinate vector of →x with respect to the basis B is CB(→x)=[a1a2]=[2−1]
Below is a video on finding a vector given basis vectors and its coordinates with respect to this basis.
Below is a video on finding the coordinates of a vector with respect to a given basis knowing its coordinates with respect to the standard basis.
Below is another video on finding a vector given basis vectors and its coordinates with respect to this basis.
Below is a video on finding a vector given basis vectors and its coordinates with respect to this basis.
Below is a video on finding the coordinates of a vector with respect to the standard basis knowing its coordinates with respect to a given basis.
Given any basis B, one can easily verify that the coordinate function is actually an isomorphism.
We now discuss the main result of this section, that is how to represent a linear transformation with respect to different bases.
Theorem 5.8.2:The Matrix of a Linear
Let T:Rn↦Rm be a linear transformation, and let B1 and B2 be bases of Rn and Rm respectively.
Then the following holds CB2T=MB2B1CB1 where MB2B1 is a unique m×n matrix.
If the basis B1 is given by B1={→v1,⋯,→vn} in this order, then MB2B1=[CB2(T(→v1))CB2(T(→v2))⋯CB2(T(→vn))]
- Proof
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The above equation (???) can be represented by the following diagram. TRn→RmCB1↓∘↓CB2Rn→RmMB2B1
Since CB1 is an isomorphism, then the matrix we are looking for is the matrix of the linear transformation CB2TC−1B1:Rn↦Rm. By Theorem 5.2.2, the columns are given by the image of the standard basis {→e1,→e2,⋯,→en}. But since C−1B1(→ei)=→vi, we readily obtain that MB2B1=[CB2TC−1B1(→e1)CB2TC−1B1(→22)⋯CB2TC−1B1(→en)]=[CB2(T(→v1))CB2(T(→v2))⋯CB2(T(→vn))] and this completes the proof.
Consider the following example.
Example 5.8.2:Matrix of a Linear
Let T:R2↦R2 be a linear transformation defined by T([ab])=[ba].
Consider the two bases B1={→v1,→v2}={[10],[−11]} and B2={[11],[1−1]}
Find the matrix MB2,B1 of T with respect to the bases B1 and B2.
Solution
By Theorem 5.8.2, the columns of MB2B1 are the coordinate vectors of T(→v1),T(→v2) with respect to B2.
Since T([10])=[01], a standard calculation yields [01]=(12)[11]+(−12)[1−1], the first column of MB2B1 is [12−12].
The second column is found in a similar way. We have T([−11])=[1−1], and with respect to B2 calculate: [1−1]=0[11]+1[1−1] Hence the second column of MB2B1 is given by [01]. We thus obtain MB2B1=[120−121]
We can verify that this is the correct matrix MB2B1 on the specific example →v=[3−1] First applying T gives T(→v)=T([3−1])=[−13] and one can compute that CB2([−13])=[1−2].
On the other hand, one compute CB1(→v) as CB1([3−1])=[2−1], and finally applying MB1B2 gives [120−121][2−1]=[1−2] as above.
We see that the same vector results from either method, as suggested by Theorem 5.8.2.
If the bases B1 and B2 are equal, say B, then we write MB instead of MBB. The following example illustrates how to compute such a matrix. Note that this is what we did earlier when we considered only B1=B2 to be the standard basis.
Example 5.8.3:Matrix of a Linear Transformation with respect to an Arbitrary Basis
Consider the basis B of R3 given by B={→v1,→v2,→v3}={[101],[111],[−110]} And let T:R3↦R3 be the linear transformation defined on B as: T[101]=[1−11],T[111]=[12−1],T[−110]=[011]
- Find the matrix MB of T relative to the basis B.
- Then find the usual matrix of T with respect to the standard basis of R3.
Solution
Equation (???) gives CBT=MBCB, and thus MB=CBTC−1B.
Now CB(→vi)=→ei, so the matrix of C−1B (with respect to the standard basis) is given by [C−1B(→e1)C−1B(→e2)C−1B(→e2)]=[11−1011110] Moreover the matrix of TC−1B is given by [TC−1B(→e1)TC−1B(→e2)TC−1B(→e2)]=[110−1211−11] Thus MB=CBTC−1B=[C−1B]−1[TC−1B]=[11−1011110]−1[110−1211−11]=[2−51−1400−21]
Consider how this works. Let →b=[b1b2b3] be an arbitrary vector in R3.
Apply C−1B to →b to get b1[101]+b2[111]+b3[−110] Apply T to this linear combination to obtain b1[1−11]+b2[12−1]+b3[011]=[b1+b2−b1+2b2+b3b1−b2+b3] Now take the matrix MB of the transformation (as found above) and multiply it by →b. [2−51−1400−21][b1b2b3]=[2b1−5b2+b3−b1+4b2−2b2+b3] Is this the coordinate vector of the above relative to the given basis? We check as follows. (2b1−5b2+b3)[101]+(−b1+4b2)[111]+(−2b2+b3)[−110] =[b1+b2−b1+2b2+b3b1−b2+b3] You see it is the same thing.
Now lets find the matrix of T with respect to the standard basis. Let A be this matrix. That is, multiplication by A is the same as doing T. Thus A[11−1011110]=[110−1211−11] Hence A=[110−1211−11][11−1011110]−1=[00123−3−3−24] Of course this is a very different matrix than the matrix of the linear transformation with respect to the non standard basis.