Change of Basis Matrix

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June 19th 2009, 07:19 PM
Fel

Change 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, 08:09 PM
malaygoel

Let $[a,b,c]=p[-1,1,1]+q[1,-1,1]+r[1,1,-1]$

giving
$a=-p+q+r$
$b= p-q+r$
$c= p+q-r$

which solves to
$p=\frac{b+c}{2}$
$q=\frac{a+c}{2}$
$r=\frac{b+a}{2}$
June 19th 2009, 08:46 PM
Fel

How is this a matrix though? I'm guessing there should be real numbers.
June 20th 2009, 05:02 AM
HallsofIvy

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 $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ f\end{bmatrix}$ , the first column of the matrix. So the first column of the "change of basis" matrix must be $\begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$ . Do the same with the second and third vectors in each basis to get the second and third columns.
June 20th 2009, 09:26 AM
Fel

Quote:

Originally Posted by HallsofIvy

So the first column of the "change of basis" matrix must be $\begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$ .

I don't understand how you got $\begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$ . How do we know what d and g are in the "change of basis" matrix?

Quote:

Originally Posted by HallsofIvy

Do the same with the second and third vectors in each basis to get the second and third columns.

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, 09:39 AM
arbolis

Quote:

Originally Posted by Fel

I don't understand how you got $\begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$

I'm like you here.
Also, if I understood well, Halls made a little mistake when he wrote $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ f\end{bmatrix}$ , I think it should be $\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ g\end{bmatrix}$ .
June 20th 2009, 10:34 AM
Plato

I am not sure that I fully understand what you need.
The matric $C = \left[ {\begin{array}{rrr} { - 1} & 1 & 1 \\ 1 & { - 1} & 1 \\ 1 & 1 & { - 1} \\ \end{array} } \right]^{ - 1} = \frac{1} {2}\left[ {\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} } \right]$ when applied to any point will the “coordinates” in the new base.

Example: $ C \cdot \left[ {\begin{array}{r} 2 \\ { - 4} \\ 6 \\ \end{array} } \right] = \left[ {\begin{array}{r} 1 \\ { - 4} \\ 1 \\ \end{array} } \right]$ so $\left\langle {2, - 4,6} \right\rangle = \left\langle { - 1,1,1} \right\rangle + 4\left\langle {1, - 1,1} \right\rangle - \left\langle {1,1, - 1} \right\rangle$
June 20th 2009, 12:08 PM
Fel

Quote:

Originally Posted by Plato

The matric $C = \left[ {\begin{array}{rrr} { - 1} & 1 & 1 \\ 1 & { - 1} & 1 \\ 1 & 1 & { - 1} \\ \end{array} } \right]^{ - 1} = \frac{1} {2}\left[ {\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} } \right]$

How did you get that inverse? I keep getting something else.
June 20th 2009, 12:19 PM
Plato

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Quote:

Originally Posted by Fel

How did you get that inverse? I keep getting something else.

I used a computer algebra system.