Matrices with respect to two bases

[Niles]

Homework Statement

I have two questions:

1) When I am asked to find the matrix with respect to two bases E and F, that is NOT the same as finding the transition-matrix going from E to F? If not, what does that matrix (with respect to E and F) represent then?

2) If I have a vector-space which is spanned by two vectors, how do I determine if a vector is part of that span?

 

[HallsofIvy]

Homework Statement

I have two questions:

1) When I am asked to find the matrix with respect to two bases E and F, that is NOT the same as finding the transition-matrix going from E to F? If not, what does that matrix (with respect to E and F) represent then?

First of all, "find the matrix with respect to two bases E and F" makes no sense! You mean "find the matrix of a linear transformation with respect to two bases E and F. The reason I specify that is that a linear transformation may be from one vector space U to a vector space V, and U and V do not necessairily even have the same dimension. That's why, in general, you must specify two bases. If your linear transformation happens to be from vector space U to itself, then you can use a single basis. There is a standard way of determining the matrix of a linear transformation with respect to two bases. First, you have to determine which is to be the basis for the domain and which the basis for the range (in terms of U->V above, which is the basis for U and which is the basis for V). If you really do have two different spaces, then it would be clear which basis belongs which vector space. If you have a linear tranformation from a vector space to itself, you must be told which basis is to be for the "domain" and which for the "range" or you can get two different matrices representing the same linear transformation. Also, in order to write a linear transformation as a matrix whether with respect to one basis or two, you must be given a specific ordering of the basis vectors.

Once you know the E is to be the basis for the domain and F the basis for the range, and have a specific ordering for each, apply the linear transformation to each of the vectors in E in turn. Write the result in terms of the basis F. The coefficients of each vector in F form a column of the matrix.

For example, suppose the linear transformation, T, from R² to R² is given by T(x, y)= (3x-y, 2x+ y) and that the bases are E= {e₁= (1, 1), e₂= (1, -1)} (in that order and for the domain) and F= {f₁=(1, 0), f₂= (2, -2))} (in that order and for the range).
T(e₁)= T(1, 1)= (3-1,2+1)= (2, 3)= 5(1, 0)-(3/2)(2, -2)= 5f₁- (3/2)f₂. The first column of the matrix is [5 -3/2].
T(e₂)= T(1, -1)= (3+ 1, 2- 1)= (4, 1)= 5(1, 0)-(1/2)(2, -2)= 5f_-1- (1/2)f₂. The second column is [5, -1/2]. The matrix corresponding to T in these bases is
\left[\begin{array}{cc}5 & 5 \\ -\frac{3}{2} & -\frac{1}{2}\end{array}\right]

2) If I have a vector-space which is spanned by two vectors, how do I determine if a vector is part of that span?

By using the definition of "span"! The span of a set of vectors is the set of all vectors that can be written as a linear combination of the vectors in that set. In particular, the vector v is in the span of {e₁, e₂} if and only if there exist numbers a and b such that v= ae₁+ be₂.

To determine if v= (1, 4) is in the span of {(3, 2), (5, -1)}, determine if there exist numbers a and b such that (1, 4)= a(3, 2)+ b(5, -1)= (3a+ 5b, 2a- b). That, of course, is the same as solving the two equations 3a+ 5b= 1, 2a- b= 4.

To determine if v= (1, 4, 1) is in the span of {(3, 2, 2), (5, -1, 0)}, determine if there exist numbers a and b such that (1, 4, 1)= a(3, 2, 2)+ b(5, -1, 0)= (3a+ 5b, 2a- b, 2a). That, of course, is the same as solving the three equations 3a+ 5b= 1, 2a- b= 4, 2a= 1.

Generally, two equations in two unknowns have a unique solution. That is because, in general (as long as they are independent), two vectors will span all of R². Generally, three equations in two unknowns do not have a solution. That is because two (independent) vectors will only span a two dimensional subspace of R² and there will be an infinite number of vectors that are NOT in that subspace.

 

[Niles]

Thank you, sir. You are indeed a professor! :-)

Have a happy new year.