# Another linear algebra problem, basis and linear transformations.

## Homework Statement

The matrix A =(1,2,3;4 5 6) defines a linear transformation T: R^3-->R^2 . Find the transformation matrix for T with respect to the basis (1,0,1),(0,2,0),(-1,0,1) for R^3 and the basis (0,1),(1,0) for R^2.

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## The Attempt at a Solution

I have no idea really. How do I change basis here when I have to basis?

The matrix you have is a representation of an abstract linear transformation RELATIVE to a specific basis. It's domain and range spaces will both have bases which are indicated by its form. A good starting point would be to examine how that transformation acts on the basis vectors of the second set.

I'm not sure I understand. What set?

I'm not sure I understand. What set?

the basis vectors form a set. {v1, v1, ..., vn} I meant how the transformation acts on {vi} you want to see how the transformation acts on the second set of basis vectors.

Ah, it transforms it. I don't see where this is leading me, however :/

Ah, it transforms it. I don't see where this is leading me, however :/

Does the question specify the representation of T is in the standard bases for R3 and R2?

yes,it does.

The the general procedure is as follows:
you are given bases, which can be though of as rules transforming a vector from one coordinate system to another. the list of basis elements you are given are the basis vectors of your space written in the standard coordinate system. If you multiply a vector in you 'alternate' coordinates by your basis vectors (as a matrix they can be the columns) you will then convert it to a vector in standard coordinates. In this case you want to take a vector in standard coordinates and convert it to a vector in alternate coordinates.

If {(1,0,1)T, (0,2,0)T, (-1,0,1)T} convert 'alternate vectors' into 'standard vectors' then the inverse of this matrix will take 'standard vectors' to alternate vectors. and likewise for the range space. The transformation matrix (given in standard coordinates) must then be transformed as well, given the inverse elements dimensions you will hopefully see the way to do this.

you have a few elements to look at: Te (that is, T in the standard basis), Peb ={(1,0,1)T, (0,2,0)T, (-1,0,1)T} (take these as the columns of a 3x3 matrix and Beb ={(0,1)T, (1,0)T}. What you want is Pbe and Bbe. Fortunately these are the inverses of Peb and Beb. Hopefully you have it from here!