Hi,
can somebody check if this is correct:
Problem:
Given two bases X_{i} and Y_{i} such that Xi=a_{ji}*Y_{i}. (1)Prove that the matrix with element a_{ij} is non singular. (2)Hence find the inverse transformation.
Solution (1)
X=A*Y
If X and Y are bases, then Vectors (x1, x2,...,xn) and (y1,y2,...yn) are linear independent. So det(X) and det(Y)≠0.
det(X)=det(AY)=det(A)*det(Y)≠0 . If det(X), det(Y)≠0 then det(A)≠0.
Conclusion: Matrix A is non singular.
Can somebody give me a hint how to find inverse transormation?
Thanks
Hi emakarov,
Yes, Xi=aji*Yj does mean X_i=\sum_{j=1}^n a_{ji}Y_j?
Xi and Yi are two sets of linear independent vectors. Elements of Xi are vectors (x1, x2, ..., xn) and element of Yi are (y1, y2, ... , yn).
Matrices X and Y are defined as nxn matrices. Matrix X contains components of x1, x2, ..., xn vectors, and matrix Y contains components of y1, y2, ..., yn vectors.
Are these assumptions correct?
Thanks
No, this is not correct.
Why are you ending your sentence with a question mark. Also, you can wrap this formula in the [tex]...[/tex] tags so that it looks better.
No. The index i of Xi is not a fixed decoration like prime '. It's a number that ranges from 1 to some n. That is, there is not one Xi, but n X's: X_{1}, X_{2}, ..., X_{n}, and similarly for Y. Each X_{i} is a vector, an element of a basis. At least this is my understanding of the problem statement.
I am not sure what a component of a vector is. As for coordinates, one can't refer to coordinates of a vector until a basis is specified. There are two bases here, but it is not clear with respect to which to consider coordinates. You can call the first basis (a sequence of vectors) X and the second one Y, but then X and Y are not matrices containing numbers; they are just sequences of vectors.
Hi emakarov,
thank you for your considerations. Sorry about question mark.
If I have understood correctly, i can write:
V^{n} is a vector space over a field F.
Two bases of V^{n} are sets X and Y.
Any vector in set X can be represented by linear combination of vectors in set Y.
.
.
.
where a_{ij} is element of F
therefore a change basis matrix "A", from X to Y is nxn matrix:
a_{11} ... a_{1n}
. .
. .
. .
a_{n1} ... a_{nn}
Similar any vector in set Y can be represented as linear combination of vectors in X.
So we can write: Y_{i}=b_{ji}*X_{j}
where B is a change matrix from Y to X.
let vector v∈V^{n}
v in respect to the basis X is equal to:
v in respect to the basis Y is equal to:
hence,
AB=I
which implies that A is a nonsingular matrix, and matrix B is the inverse of A