How do I solve for a matrix using base vectors and vector spaces?

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SUMMARY

The discussion centers on solving for a matrix A that transforms base vectors e1=(1,-2,0) and e2=(0,3,0) into vectors w1=(1,0,0) and w2=(0,1,0). The initial calculation of matrix A as (1, 0, 0; -2, 3, 0; 0, 0, 0) is incorrect due to the non-existence of its inverse. The correct approach involves recognizing that A must be reduced to a two-dimensional space, leading to the correct inverse matrix A as (1,0;-2,3).

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Kruger
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Homework Statement



The problem is quite easy, but I've still trouble solving this.

Given the two base vectors e1=(1,-2,0) and e2=(0,3,0) and the other ones of a different vector space w1=(1,0,0) and w2=(0,1,0).

I've to find a matrix A that that does the following Ae1=w1 and Ae2=w2

2. The attempt at a solution

Easy isn't it? I've done what the professor did to solve such problems:

Calculate: e1=1*w1+(-2)*w2
and: e2=0*w1+3*w2

thus that shouls yield the matrix A:(1, 0, 0; -2, 3, 0; 0, 0, 0)
where ; is written for different lines in the matrix A.

But if I calculate A*e1 I get something totally wrong.

Where's the mistake in my calculation?
 
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You calculated the inverse of A, expressing the e vectors as a linear combination of the w vectors:

Ae1=w1 so e1=A^(-1)w1
Ae2=w2 so e2=A^(-1)w2

Clearly, your answer for A is incorrect since the inverse of your A does NOT exist. For this reason, you need to drop the third dimension. So , for example, e1 becomes (1,-2) etc.

The way you proceed is correct though, except that your A is actually the inverse of A, and the inverse of A is indeed (1,0;-2,3). So acquire A now.

marlon
 
Last edited:
I don't understand exactly why I've to build the inverse of A, because I search A such that A*e1=w1 not that A*w1=e1.
 
ahh, k, I got it know, I took your hint marlon, thanks, thanks.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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