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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- Jun 19th 2009, 07:19 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! - Jun 19th 2009, 08:09 PMmalaygoel
Let$\displaystyle [a,b,c]=p[-1,1,1]+q[1,-1,1]+r[1,1,-1]$

giving

$\displaystyle a=-p+q+r$

$\displaystyle b= p-q+r$

$\displaystyle c= p+q-r$

which solves to

$\displaystyle p=\frac{b+c}{2}$

$\displaystyle q=\frac{a+c}{2}$

$\displaystyle r=\frac{b+a}{2}$ - Jun 19th 2009, 08:46 PMFel
How is this a matrix though? I'm guessing there should be real numbers.

- Jun 20th 2009, 05:02 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 $\displaystyle \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ f\end{bmatrix}$, the first column of the matrix. So the first column of the "change of basis" matrix must be $\displaystyle \begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$. Do the same with the second and third vectors in each basis to get the second and third columns. - Jun 20th 2009, 09:26 AMFel
I don't understand how you got $\displaystyle \begin{bmatrix}0 \\\frac{1}{2}\\ \frac{1}{2}\end{bmatrix}$. How do we know what d and g are in the "change of basis" matrix?

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. - Jun 20th 2009, 09:39 AMarbolis
I'm like you here.

Also, if I understood well, Halls made a little mistake when he wrote $\displaystyle \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ f\end{bmatrix}$, I think it should be $\displaystyle \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ d\\ g\end{bmatrix}$. - Jun 20th 2009, 10:34 AMPlato
I am not sure that I fully understand what you need.

The matric $\displaystyle 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: $\displaystyle

C \cdot \left[ {\begin{array}{r}

2 \\

{ - 4} \\

6 \\

\end{array} } \right] = \left[ {\begin{array}{r}

1 \\

{ - 4} \\

1 \\ \end{array} } \right]$ so $\displaystyle \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 $ - Jun 20th 2009, 12:08 PMFel
- Jun 20th 2009, 12:19 PMPlato