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Consider the basis $B=\left\{\begin{pmatrix} -1 \\ 1 \\0 \end{pmatrix}\begin{pmatrix} -1 \\ 0 \\1 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\1 \end{pmatrix} \right\}$ for $\mathbb{R}^3$.

A) Find the change of basis matrix for converting from the standard basis to the basis B.

I have never done anything like this and the only examples I can find online basically tell me how to do the change of basis for "change-of-coordinates matrix from B to C".

B) Write the vector $\begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}$ in B-coordinates.

Obviously I cant do this if I cant complete part A.

Can someone either give me a hint, or preferably guide me towards an example of this type of problem?

The absolute only thing I can think to do is take an augmented matrix $[B E]$ (note - E in this case is the standard basis, because I don't know the correct notation) and row reduce until B is now the standard matrix. This is basically finding the inverse, so I doubt this is correct. Any help would be much appreciated.

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In which basis is your B defined? I see that it is a collection of 3 vectors $B = [\vec B_1 \vec B_2 \vec B_3]$ whereas each of the $B_n$ is a vector of coordinates. Coordinates must be specified wrt to some another basis (or with B itself?). What is that basis? – Val Jun 3 '14 at 11:59

2 Answers 2

up vote 12 down vote accepted

Denote $E$ the canonical basis of $\mathbb{R}^3$.

A) These three column vectors define a $3\times 3$ matrix $$P=\left(\matrix{-1&-1&1\\1&0&1\\0&1&1}\right)$$ which is the matrix of the linear map $$ Id:(\mathbb{R}^3,B)\longrightarrow (\mathbb{R}^3,E). $$ This means in particular that whenever you right multiply it by a column vector $(x_1,x_2,x_3)$ where $x_j$ are the coordinates of a vector $x=x_1B_1+x_2B_2+x_3B_3$ with the respect to the basis $B$, you obtain the coordinates of $x$ in the canonical basis $E$.

What you want is the matrix of $$ Id:(\mathbb{R}^3,E)\longrightarrow (\mathbb{R}^3,B). $$ That is $P^{-1}$, the inverse of the matrix above. This will transform, by right multiplication, the coordinates of a vector with respect to $E$ into its coordinates with respect to $B$. That's the change of basis matrix you need.

B) As explained above, you just have to right multiply the change of basis matrix $P^{-1}$ by this column vector.

Check your answer: you should find

$$P^{-1}=\left(\matrix{-1/3&2/3&-1/3\\-1/3&-1/3&2/3\\1/3&1/3&1/3} \right)$$ $$\left(\matrix{-1/3&2/3&-1/3\\-1/3&-1/3&2/3\\1/3&1/3&1/3} \right)\left(\matrix{1\\0\\0}\right)=\left(\matrix{-1/3\\-1/3\\1/3}\right).$$

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Can someone please show me how to use the spoiler option? Thanks! – 1015 Mar 25 '13 at 20:06
It seems that I'm doing exactly like they show here and it still does not work. I must be dumb... – 1015 Mar 25 '13 at 20:12
I believe that you need to move the closing double-dollar onto the same line as the rest of the expression for each of the two lines. I'm not sure if it'll try to merge the two lines of maths as it does (at least for me) in the preview. – Glen O Mar 25 '13 at 21:45
@GlenO Brilliant! Thanks a lot! – 1015 Mar 25 '13 at 21:49
What is Id? Is it identity matrix? If so, how can it convert anything? – Val Jun 3 '14 at 11:54

Just to clarify 1015 answer for myself

We have

$$B = [\vec b_1 \vec b_2 \vec b_3] = E \left[\matrix{-1&-1&1\\1&0&1\\0&1&1}\right] = E [B]_E = EP$$

It says that $P = [B]_E$ consists of columns of $b_n$, the basis vectors $b_n$ in basis standard $E = [\vec e_1, \vec e_2, \vec e_3]$, so that

$$\vec b_1 = [\vec e_1 \vec e_2 \vec e_3] \left[\matrix{-1\\1\\0}\right].$$

Now, we can represent any vector in basis E as well in basis B

$$\vec v = E [\vec v]_E = B [\vec v]_B = E P [\vec v]_B$$


$$[\vec v]_E = P [\vec v]_B$$

We see that P translates vector B-coordinates into E-coordinates.

In problem A), we have P, coordinates $[\vec v]_E$ of vector $\vec v$ basis E, and wish to compute them into $[\vec v]_B$. That is easy from the last equation,

$$[\vec v]_B = P^{-1}[\vec v]_E.$$

You see, $P^{-1}$ does the conversion. I call it inverse of change of basis matrix. 1015 has already computed it for your convenience. I just wanted to explain why.

For the problem B), just plug $[\vec v]_E = \left[\matrix{1\\0\\0}\right].$ I assume the standard basis, though I want to know why. Similarly, I want to know why don't you specify the basis for the components of B.

It must be noted though that textbooks normally have $$\vec v = E[\vec v]_e = EPP^{-1}[\vec v]_e = B [\vec v]_b$$ so that basis is translated by right-multiplying with change of basis matrix $P$, $$B = EP,$$ and coordinates are translated contravariantly, $[\vec v]_b = P^{-1} [\vec v]_e$.

For some reason 1015 has chosen the inverse $P^{-1}$, used to translate the coordinates, to be the change of basis matrix.

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