Change of basis matrix to convert standard basis to another basis

Question

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 can't do this if I can't complete part A.

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

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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.

Answer 1

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

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

Answer 2

By definition change of base matrix contains the coordinates of the new base in respect to old base as it's columns. So by definition $B$ is the change of base matrix. Key to solution is equation $v = Bv'$ where $v$ has coordinates in old basis and $v'$ has coordinates in the new basis (new basis is B-s cols) suppose we know that in old basis $v$ has coords $(1,0,0)$ (as a column) (which is by the way just an old base vector) and we want to know $v'$ (the old base vector coordinates in terms of new base) then from the above equation we get $$B^{-1}v = B^{-1}Bv' \Rightarrow B^{-1}v = v'$$

As a side-node, sometimes we want to ask how does that change of base matrix B act if we look at it as linear transformation, that is given vector v in old base $v=(v_1,...,v_n)$, what is the vector $Bv$? In general it is a vector whith i-th coordinate bi1*v1+...+bin*vn (dot product of i-th row of $B$ with $v$). But in particular if we consider v to be an old base vector having coordinates (0...1...0) (coordinates in respect the old base) where 1 is in the j-th position, then we get $Bv = (b_{1j},...,b_{nj})$ which is the j-th column of B, which is the j-th base vector of the new base. Thus we may say that B viewed as linear transformation takes old base to new base.

Answer 3

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

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

or

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

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.

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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.