I understood the process of finding the change of basis matrix for converting a vector $a$'s representation with respect to an arbitrary basis $B = \{ v_1, v_2, ... v_k \}$, lets call this representation $[a]_B$, to the standard representation. The change of basis matrix is simply

$$C = \begin{pmatrix} v_1 & v_2 & \cdots & v_k \end{pmatrix} $$

where $v_i$ for $i=1...k$ are in general $n$-dimensional vectors (I think). The operation is instead

$$C \cdot [a]_B = a$$

where $a$ is the standard representation of the vector. Why does this work? Basically, if $a$ is in the span of $B$, then $a$ can be represented as follows

$$a = c_1 v_1 + c_2v_2 + \cdots + c_k v_k$$

which is equivalent to

$$a = C \begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_k \end{pmatrix} = C\cdot [a]_B$$

I can also find $a$ if I know $[a]_B$, but $C$ must be invertible, as follows

$$[a]_B = C^{-1} a$$

My questions are:

  1. What if $C$ is not invertible? How can we find the change of basis matrix from one basis to the other?

  2. What about when we have two basis and neither one nor the other is the standard basis?

  3. What are the dimensions of $v_i$s?

  • $\begingroup$ A square matrix is invertible if and only if its columns are a basis. So $\endgroup$ – user228113 Feb 27 '16 at 12:28
  1. If $\mathcal B$ is a basis, $C$ is necessarily invertible, since you can express each vector of the standard basis in the new basis.
  2. The proof for the change of basis formula does not depend on using the standard basis or not. ‘Standard basis’ is a pure convention.
  3. What do you mean by ‘dimension(s)’ of a vector?
  • $\begingroup$ For the first point. A basis for $\mathbb{R}^2$ could consist of two vectors of 3 coordinates. In this case, we would not have a square change of basis matrix. $\endgroup$ – nbro Feb 27 '16 at 12:56
  • $\begingroup$ How can a vector have $3$ coordinates in $\mathbf R^{\color{red}2}$? $\endgroup$ – Bernard Feb 27 '16 at 13:03
  • $\begingroup$ I have here an example of a basis with two 3-dimensional vectors. Then vectors with respect to this basis only have two coordinates, but this doesn't change anything regarding the dimension of the vectors in the basis. I also have that the change of basis matrix in on direction would exist (and you can construct it as I'm describing above), but of course the inverse would not exist, since the change of basis matrix is not even squared. $\endgroup$ – nbro Feb 27 '16 at 13:25
  • $\begingroup$ Regarding your second point. I think you're not right. It matters. Usually the coordinates of a basis are given with respect to the standard basis. If we have two general basis $B$ and $A$ that are not equal to the standard basis (of some dimension), the vectors in $B$ and $A$ are usually represented with respect to the standard basis (unless otherwise specified, which I've not seen yet). So, if you want to find the change of basis matrix from $A$ to $B$ you can't simply create the matrix whose columns are the vectors in $A$. <cont> $\endgroup$ – nbro Feb 27 '16 at 13:30
  • $\begingroup$ <cont>We probably need first to find the change of basis from $A$ to the standard basis, and then from the standard basis to $B$. $\endgroup$ – nbro Feb 27 '16 at 13:30

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