The Matrix Of A Linear Transformation

[playboy]

I saw a similar post to this one, but i just got lost in the mess of the whole thing. So i just started a new thread.

A question reads:

Let T: Pn ---> Pn be defined by T[P(x)] = p(x) + xp'(x), where p'(x) denotes the derivative. Show that T is an isomorphism by finding Mbb(T) when B = {1, x, x^2, ... , x^n}

From doing the other question and problems in the textbook, i know how to find Mdb(T). I suppose that finding Mbb(T) would be very similar.

I did it like this:

Mbb(T) = [ CbT(1) CbT(x) CbT(x^2) ... CbT(X^n) ]
Mbb(T) = [ Cb(1) Cb(2x) Cb(3x^2) ... Cb((n+1)X^n]

and it gives this nxn matrix:

[1 0 0 ... 0]
[0 2 0 ... 0]
[0 0 3 ... 0]
[0 0 0 ... (n+1)]

now, an isomorphism means that the linear transformation is both one-to-one and onto.

How do you tell that its an isomorphism by just looking at the matrix?

 

[Physics Monkey]

In this case, your transformation is isomorphic if the matrix is invertible. Why?

 

[playboy]

Their is a theorem in my textbook.

"Let A be an mXn matrix, and let TA : R^n ---> R^m be the linear transformation inudced by A, that is, Ta(X) = AX for all X in R^n.

1. Ta is onto if and only if rank A = m
2. Ta is one-to-one if and only if A = n

oh wait.. that dosn't work in this problem...

ill brb

 

[playboy]

Their is an example in the textbook...

If U is any invertable mxm matrix, the map T: Mmn ----> Mmn given by T(X) = UX is an isomorphism by Example 6 Section 8.2

This example is too long to type out... but it dosn't show any matrix. It just uses the transformation to show it.