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Change of basis using RowReduce

Posted 7 months ago
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I'm trying to set up a change of basis matrix from the base W -> V and then verify that it works by seeing that the co-ordinates on the x-vector changes as they should. I'm supposed to use the Inverse function, but I don't know what to do with it. I could really appreciate some help. This is what I've got so far:

V = {{1, 3}, {4, 6}}
W = {{4, 6}, {2, 5}} 

x = {6, 6}

So, my question is, how do I do I setup the change of basis matrix from W -> V, and then verify it?

Thanks in advance.

3 Replies

Why don't you begin by describing what one would do so as to obtain the change-of-basis matrix (before necessarily involving Mathematica): express each vector of the new basis as a linear combination of the vectors in the old basis.

"just ignore the new wording"

You can say T(x) = Ax

If T goes from R^n to R^m, (let''s say m>n). see this example of n=2 m=3 and finding A.

T(e1)={5,-7,2}, T(e2)={-3,8,0}, x={x1,x2}, = x1e1+x2e2 and T(x)=x1[T(e1)+x2T(e2) or ...





If you want it "invertible" or reverse solved, A must be an invertible matrix (made of elemetary matrices or row operations, meaning "it can solve" without free variable or other issues).

You shouldn't be mystified by it. You are just multiplying, say a A a 2x3 matrix by a, say, x is a 1x2 and by rule ending up with a 3x3. the 3x3 may be "onto" or "one to one" (which depends on A)

Ax=T(x) = blah, a 3x3 matrix. simple multiplication.

Now, A A^-1 = I, A^-1 A = I, and [A I] row reduced exposes x (one way to find inverse), and there are a list of rules telling you if a matrix will be invertible (which you should review). If BA=I or AB=I, then B=A^-1 or A=B^-1. Also there is the topic of "a matrix composed of elementary matrices" which you should remember because you'll be needing it like a carpenter would a level.

So if your asked:



you have Ax=b (b={3,7}) and x=A^-1 b, you solve as you learned in earlier lessons

your new situation is no different: just re-worded

S(T(x))=S(Ax)=A^-1 (Ax)=x

"just ignore the new wording"

Back to your problem you have V, W, x they are all constant. I might say "find T for V->W so that Vx->W" but i wouldn't since you have all these defined as constant. That means I'm unsure of what the problems is asking.

"problem theory": ALWAYS ask if the problem is asking for a number as an answer or to change and equation from one form to another form as an answer. Do not try to solve a problem if the terms are unclear. You are given data but (as i am sitting) i have not been asked to do anything. So i should do nothing but demand i be told what it is I am being asked to do :) I could be getting asked for Wxb=V and what is b, but i cannot say i am. I could be getting asked WV=xC, do i know?

as far as invertible, i think i covered that above

another way of unraveling confusion is this:

Ax=b simply means multiply A and x, multiplied using "matrix multiplication" equals b. A can represent a set of equations, but that is only one use.

Finding an inverse simply means the same as it does in algebra: but the process of finding the inverse is detoured by that multiplications were "ordered".

always: the zero property is used, as you learned in algebra. there are just more "steps" to take.

x+4=5, what is x? the zero property. but take a matrix and ordered multiplication then you need to apply the zero property multiply to "undo" things. the zero property is the row reduction rules you learned earlier - they all simply obey the simple algebra property.

it can be unwieldy and costly for a student to over-think a matrix as being an underlying system of equations behind every problem with a determinant or other solution having graphical implications and etc.

a matrix multiply can mean a million things: there is no end to what it might mean. so don't try to understand why it is used in which way. just that it works in the way currently being discussed.

i suggest you think "just ignore the new wording", learn the re-wording, move on to next chapter

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