# Linear transformations, vectors

• Oct 29th 2013, 04:01 AM
FilipVz
Linear transformations, vectors
Hi,

can somebody check if this is correct:

Problem:

Given two bases Xi and Yi such that Xi=aji*Yi. (1)Prove that the matrix with element aij 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 :)
• Oct 29th 2013, 08:12 AM
emakarov
Re: Linear transformations, vectors
Quote:

Originally Posted by FilipVz
Given two bases Xi and Yi such that Xi=aji*Yi.

Do you mean Xi=aji*Yj? 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?
• Oct 29th 2013, 08:25 AM
FilipVz
Re: Linear transformations, vectors
X and y are nxn matrices.
• Oct 29th 2013, 08:34 AM
emakarov
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!
• Oct 29th 2013, 12:20 PM
FilipVz
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
• Oct 29th 2013, 01:35 PM
emakarov
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 $$...$$ 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: X1, X2, ..., Xn, and similarly for Y. Each Xi 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.
• Nov 2nd 2013, 01:08 PM
FilipVz
Re: Linear transformations, vectors
Hi emakarov,

If I have understood correctly, i can write:

Vn is a vector space over a field F.

Two bases of Vn 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 aij is element of F

therefore a change basis matrix "A", from X to Y is nxn matrix:

a11 ... a1n
. .
. .
. .
an1 ... ann

Similar any vector in set Y can be represented as linear combination of vectors in X.
So we can write: Yi=bji*Xj

where B is a change matrix from Y to X.

let vector v∈Vn

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