# Linear operators and change of basis in a vector space

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).

• Why do you write $\;a_i\neq b_i\;$ ? Is that a condition on the basis and on some specific $\;v\in V\;$ ? Sep 4, 2014 at 19:24
• 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.
– Will
Sep 4, 2014 at 19:28

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.)

• 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?
– Will
Sep 4, 2014 at 19:43
• 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). Sep 4, 2014 at 19:51
• 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 $\mathcal{E}$.
– Will
Sep 4, 2014 at 20:02