# Change of basis matrix to convert standard basis to another basis

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?

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.

• 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, 2014 at 11:59

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.

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

• What is Id? Is it identity matrix? If so, how can it convert anything?
– Val
Jun 3, 2014 at 11:54
• It's the name of the function I assume Jun 4, 2014 at 13:07
• Id is an identity mapping, since in general there are three steps: (1) represent the input vector in the input basis, (2) do a mapping, which in this case is the idneity mapping, ie do nothing, and then (3) represent the result in the output basis. Feb 5, 2017 at 17:58

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.

• As another side-note, if both old basis and new basis are orthonormal then finding inverse of B is simple: inverse(B) = transpose(B), but this is not applicable to the B in the original post, because it is not orthonormal. Mar 27, 2018 at 12:10

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.

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.