Hey guys, I have a question on change of basis.

Find the change of basis matrix from the standard basis {(1,0,0),(0,1,0),(0,0,1)} to {(-1,1,1),(1,-1,1),(1,1,-1)}.

I'm having some trouble in figuring out what to do. Thanks in advance!

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- June 19th 2009, 08:19 PMFelChange of Basis Matrix
Hey guys, I have a question on change of basis.

Find the change of basis matrix from the standard basis {(1,0,0),(0,1,0),(0,0,1)} to {(-1,1,1),(1,-1,1),(1,1,-1)}.

I'm having some trouble in figuring out what to do. Thanks in advance! - June 19th 2009, 09:09 PMmalaygoel
Let

giving

which solves to

- June 19th 2009, 09:46 PMFel
How is this a matrix though? I'm guessing there should be real numbers.

- June 20th 2009, 06:02 AMHallsofIvy
The "change of basis" matrix must map the coefficients of a given vector, as written in one basis, into the coefficients of the same vector, written in the other basis.

For example, the vector [1, 0, 0], in the first basis would be written, of course, as 1[1, 0, 0]+ 0[0, 1, 0]+ 0[0, 0, 1] so its coefficents are [1, 0, 0] (which is why it is the 'standard' basis). In terms of the other basis, [1, 0, 0]= a[-1, 1, 1]+ b[1, -1, 1]+ c[1, 1, -1]= [-a+b+c, a-b+c, a+b-c]. So we must have -a+ b+ c= 1, a- b+ c= 0, a+ b- c= 0. Adding the first two equatins, 2c= 1 so c= 1/2. Adding the last two equations, 2a= 0 so a= 0, Adding the first and last equations, 2b= 1 so b= 1/2. The coefficients are [0, 1/2, 1/2].

But it should be easy to see that if we multiply any matrix by [1, 0, 0] we get , the first column of the matrix. So the first column of the "change of basis" matrix must be . Do the same with the second and third vectors in each basis to get the second and third columns. - June 20th 2009, 10:26 AMFel
I don't understand how you got . How do we know what d and g are in the "change of basis" matrix?

Shouldn't the first vector that you multiply by be (-1,1,1) and not (1,0,0)? And then the second and third vectors would be (1,-1,1) and (1,1,-1) respectively? Just a bit confused but thanks for taking the time to help. - June 20th 2009, 10:39 AMarbolis
- June 20th 2009, 11:34 AMPlato
I am not sure that I fully understand what you need.

The matric when applied to any point will the “coordinates” in the new base.

Example: so - June 20th 2009, 01:08 PMFel
- June 20th 2009, 01:19 PMPlato