Linear transformations, vectors

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 :)

Re: Linear transformations, vectors

Quote:

Originally Posted by

**FilipVz** Given two bases X_{i} and Y_{i} such that Xi=a_{ji}*Y_{i}.

Do you mean Xi=a_{ji}*Y_{j}? And does this mean $\displaystyle X_i=\sum_{j=1}^n a_{ji}Y_j$?

Quote:

Originally Posted by

**FilipVz** X=A*Y

What do you mean by this? What exactly are X and Y here?

Re: Linear transformations, vectors

X and y are nxn matrices.

Re: Linear transformations, vectors

Quote:

Originally Posted by

**FilipVz** X and y are nxn matrices.

Matrices containing **what**? You start with some vectors $\displaystyle X_i$ and $\displaystyle Y_j$; there is no talk about numbers! And please don't say "Coordinates of $\displaystyle X_i$" and stop there!

Re: Linear transformations, vectors

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

Re: Linear transformations, vectors

No, this is not correct.

Quote:

Originally Posted by

**FilipVz** Yes, Xi=aji*Yj does mean X_i=\sum_{j=1}^n a_{ji}Y_j?

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.

Quote:

Originally Posted by

**FilipVz** 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).

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.

Quote:

Originally Posted by

**FilipVz** 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.

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.

Re: Linear transformations, 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.

$\displaystyle X={(X_1,X_2,…,X_n)}$

$\displaystyle Y={(Y_1,Y,…,Y_n)} $

Any vector in set X can be represented by linear combination of vectors in set Y.

$\displaystyle X_1=a_11 Y_1+ …+a_1n Y_n$

.

.

.

$\displaystyle X_n=a_n1 Y_1+ …+a_nn Y_n$

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:

$\displaystyle [v]_X=A[v]_Y$

**v** in respect to the basis Y is equal to:

$\displaystyle [v]_Y=B[v]_X$

hence,

$\displaystyle [v]_X=AB[v]_X$

AB=I

which implies that A is a nonsingular matrix, and matrix B is the inverse of A