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.

  • $\begingroup$ 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? $\endgroup$ – Val Jun 3 '14 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.

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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    $\begingroup$ Can someone please show me how to use the spoiler option? Thanks! $\endgroup$ – Julien Mar 25 '13 at 20:06
  • $\begingroup$ It seems that I'm doing exactly like they show here meta.stackoverflow.com/posts/71396/edit and it still does not work. I must be dumb... $\endgroup$ – Julien Mar 25 '13 at 20:12
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    $\begingroup$ What is Id? Is it identity matrix? If so, how can it convert anything? $\endgroup$ – Val Jun 3 '14 at 11:54
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    $\begingroup$ It's the name of the function I assume $\endgroup$ – TheWaveLad Jun 4 '14 at 13:07
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    $\begingroup$ 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. $\endgroup$ – Hugh Perkins Feb 5 '17 at 17:58

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.


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 = B*v' 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 inverse(B)*v = inverse(B)Bv' => inverse(B)*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=(v1,...,vn), what is the vector B*v? In general it is a vector whith i-th coordinate bi1*v1+...+binvn (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 = (b1j,...,bnj) 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.

  • $\begingroup$ 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. $\endgroup$ – Jaanus Kiipli Mar 27 '18 at 12:10

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