Suppose we have a vector space $V$ over a scalar field $\mathbb{F}$ and two different bases $\mathcal{B}=\lbrace\mathbf{v}_{i}\rbrace_{i=1,\ldots , n}$ and $\mathcal{E}=\lbrace\mathbf{w}_{i}\rbrace_{i=1,\ldots , n}$. A given vector $\mathbf{x} \in V$ can be described in terms of either of these bases, by the (unique) linear combinations $$\mathbf{x}= \sum_{i=1}^{n}a_{i}\mathbf{v}_{i} \quad \text{or}\quad \mathbf{x}= \sum_{i=1}^{n}b_{i}\mathbf{w}_{i}$$ where $a_{i} \neq b_{i}$.

Now, suppose that we have a linear operator $\mathcal{S}$ which maps the basis $\mathcal{E}$ to the basis $\mathcal{B}$, defined in the following manner, $$\mathcal{S}\left(\mathbf{w}_{j}\right)= \mathbf{v}_{j}$$ As $\mathbf{v}_{j} \in V$ it can itself be described in terms of a (unique) linear combination of the basis vectors $\mathbf{w}_{i}$ $\left(i=1,\ldots , n\right)$ and as such, we have that $$ \mathbf{v}_{j}=\sum_{i=1}^{n}\left[\mathcal{S}\left(\mathbf{w}_{j}\right)\right]_{i}\mathbf{w}_{i}=\sum_{i=1}^{n}S_{ij}\mathbf{w}_{i}$$ where we have defined $\left[\mathcal{S}\left(\mathbf{w}_{j}\right)\right]_{i} \equiv S_{ij}$, which is the $i^{th}$ component of the $j^{th}$ basis vector $\mathbf{v}_{j} \in \mathcal{B}$ with respect to the basis $\mathcal{E}$.

Is this a correct description of how to change between two different bases for the same vector space? (The equation $\mathbf{v}_{j}=\sum_{i=1}^{n}S_{ij}\mathbf{w}_{i}$ is given in a text book that I've been reading as a formula for switching between two different basis sets for a given vector space and I'm trying to justify its form).

  • $\begingroup$ Why do you write $\;a_i\neq b_i\;$ ? Is that a condition on the basis and on some specific $\;v\in V\;$ ? $\endgroup$
    – Timbuc
    Sep 4, 2014 at 19:24
  • $\begingroup$ Sorry, it's a little ambiguous, by $a_{i} \neq b_{i}$ I just meant that the components of a given vector with respect to the two given bases (respectively) will not, in general, be equal to one another. $\endgroup$
    – Will
    Sep 4, 2014 at 19:28

1 Answer 1


To see the pattern it helps to write out specific cases. Such as $$v_1=a_1 w_1 + a_2 w_2 + a_3 w_3 + ...$$ and $$v_2=b_1 w_1 + b_2 w_2 + b_3 w_3 + ...$$ and $$v_3 = c_1 w_1 + c_3 w_3 + ...$$

Now you're really tired of the alphabet game and aching for something more concise, so you think maybe $a$ can be denoted by $S_1$ and $b$ by $S_2$ and ... etc. And you're into double subscripts. Note that the subscript of $S$ matches up with the subscript of $v$ and the subscript that increments through each sum matches up with $w$. Then you've understand the equation your author used.

Now look back at what you originally wrote:$$\sum_{i=1}^{n}\left[\mathcal{S}\left(\mathbf{w}_{j}\right)\right]_{i}\mathbf{w}_{i}$$

That glob with the brackets is some kind of private notation of yours unknown to the rest of us. In my alphabet-infested equations, if I replace $a_1$ by $S$ with 1,1 subscript and $a_2$ by $S$ with 1,2 subscript and $a_3$ by $S$ with 1,3 subscript, and if I use $i$ for the varying subscript in the sum, then I should use $j$ for the subscript of the $v$'s on the left hand side of my equations. If you rewrite my alphabet-infested equations with that convention, I think you'll get what your author wrote, or the equivalent. (Which subscript you decide to designate by $i$ and which by $j$ obviously doesn't matter. If that's not obvious you'll have to ponder it and fool around with concrete examples of double-subscripting until it is.)

  • $\begingroup$ Thanks. Yeah, I understand that part of it, but was really trying to describe the concept in terms of a linear operator acting on the basis vectors. Would the description I've given be correct with respect to that? $\endgroup$
    – Will
    Sep 4, 2014 at 19:43
  • $\begingroup$ Everything in your original write-up makes sense except for the funky bracket glob -- which isn't consistent with the normal way the math is written, and so forces your reader to guess your meaning (not what you want to do). $\endgroup$
    – kmiker
    Sep 4, 2014 at 19:51
  • $\begingroup$ Ah ok, thanks. Yes, sorry it's a bit messy. The notation was an attempt to make it explicit that it was the $i^{th}$ component of that vector $\mathbf{v}_{j}$ with respect to the other basis $\mathcal{E}$. $\endgroup$
    – Will
    Sep 4, 2014 at 20:02

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