# Change of Basis and Row Operations

I want to find the shortest way of expressing the following idea so that I can use it as a lemma during exams.

I want to execute a change of basis without having to go through the whole process of calculating the change of basis matrix $M$, and then applying it by performing $C = MAM^{-1}$.

Given a matrix $A_{n \times n}$ and bases $B$ and $B'$ according to their coordinate vectors in base $E$ and assuming that $A$ is currently expressed according to $B$, I should be able to take the expanded matrix:

\begin{bmatrix} [B]_E & | & [A]_B \end{bmatrix}

and perform a series of row operations on it to arrive at:

\begin{bmatrix} [B']_E & | & [A]_{B'} \end{bmatrix}

the idea being similar to calculating the inverse by going from

\begin{bmatrix} I & | & A \end{bmatrix} to \begin{bmatrix} A^{-1} & | & I \end{bmatrix}

The first thing is whether or not this is correct. I think it is since any matrix / linear transformation can be decomposed into a series of consecutive elementary ( matrices / row ) operations.

If this is correct, what's the shortest acceptable way of stating this?

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Note that row operations can only change the range basis. You need to perform column operations to change the domain basis. –  copper.hat Nov 4 '12 at 10:06

It is not correct, basically because the change of basis operation that transforms $A$ into $MAM^{-1}$ involves both left-multiplication (by $M$) and right-multiplication (by $M^{-1}$); the former is realised by row operations, while the latter is realised by column operations, so you need both.

There is also a fundamental problem that bases are not expressed by matrices, they just are collections of vectors. Only if you've got (yet) another basis $E$ can you represent a basis $B$ by expressing its vectors in the basis $E$, and list them as columns of a matrix.

If you really must proceed by incremental changes of basis (I don't know of any practical applications of this), this is possible by repeatedly applying the change of basis formula. You should be well aware of what you do there though: the matrix $M$ as you gave it transforms the coordinates on the old basis into coordinates on the new basis, it does not transform the old basis itself. Thus when you apply $C=MAM^{-1}$ to a vector expressed on the new basis, $M^{-1}$ first expresses it on the old basis so that $A$ can act on them, and finally $M$ converts the resulting (old) coordinates into ones on the new basis again. The new basis itself expressed on the old basis is given by the columns of $M^{-1}$: these are obtained by converting the coordinates of the new basis on itself (which are the columns of the identity matrix) into coordinates on the old basis by applying $M^{-1}$.

So if you want to change from a basis $B$ to one differing slightly from $B$, the matrix that expresses the change takes the place of $M^{-1}$. Concretely let's say $B=[b_1,b_2,b_3]$ is changed into $B'=[b1,b_2+7b_3,b_3]$, which can be expressed by the matrix $$P=\begin{pmatrix}1&0&0\\0&1&0\\0&7&1\\\end{pmatrix},$$ then $P$ takes the role of $M^{-1}$. So to change what you call $[A]_B$ into $[A]_{B'}$ you should right-multiply by $P$ (which means add the third column $7$ times to the second column) and also left-multiply by $P^{-1}$, which has the entry $7$ replaced by $-7$ (this means subtract the second row $7$ times from the third row).

Note that what we did to pass from $B$ to $B'$ corresponds to just right-multiplication by $P$ (add the thrid vector $7$ times to the second). So you could perform something like what you proposed by only performing column operations to the basis (adding some vectors to other vectors), and every time you do so, perform the same column operation to the matrix of you linear map and then immediately also perform the "inverse" row operation to it. Be sure you understand precisely what this inverse row operation means in all cases, or you'll go astray. I think you'd do better to stay with the direct change of basis formula though, and compute the inverse matrix needed there as you indicated (doing so be sure to use only row operations!).

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That's the idea, the bases $B$ and $B'$ are expressed in terms of $E$. I.E. the matrices $B$ and $B'$ are meant to be the set of coordinate vectors according to $E$. I'll edit that to make it clear. –  Robert S. Barnes Nov 4 '12 at 8:26
I see my problem now. $B$ is expressed in terms of $E$, but $A$ is expressed in terms of $B$. For this to work $B$ I have to take the coordinated vectors of $B$ according to $B$. Then I could use row operations to go from \begin{bmatrix} [B]_B & | & [A]_B \end{bmatrix} to \begin{bmatrix} [B]_{B'} & | & [A]_{B'} \end{bmatrix} Does that sound correct? The only easy case would then be if I'm going from \begin{bmatrix} E & | & [A]_E \end{bmatrix} to \begin{bmatrix} [E]_B & | & [A]_B \end{bmatrix} –  Robert S. Barnes Nov 4 '12 at 8:43
I posted my comment just as you posted your edit, so I'm not sure yet if my comment is correct according to what you said in your edit. –  Robert S. Barnes Nov 4 '12 at 8:50

This is not an answer, but may help. It really belongs as a comment, but is too long. It is a summary of notation I found useful in Knapp's Basic Algebra (it may be more widespread, but this was the only place I encountered it). Everything here is finite dimensional.

There is nothing startling here, just a notational aid.

A basis is represented as an ordered tuple. So $B=(b_1,...,b_n)$, etc.

An element $u\in U$ where $U$ is a vector space has a representation $u=\sum_k x_k b_k$, where $b_k$ is an ordered basis for $U$. Knapp's notation is $x = \binom{u}{B}$, meaning that $x$ is the vector representation of $u$ in the (ordered) basis $B$.

An operator $L:U \to U$, with ordered bases $B$ and $C=(c_1,...,c_n)$ for domain and range respectively, has a matrix representation $A$, defined by $L b_i = \sum_{i,j} [A]_{ij} c_j$. Knapp's notation is $A = \binom{L}{C \ B}$, meaning that $A$ is the matrix representation of $L$ with basis $B$ for input and $C$ for output.

The point is that these notations 'work nicely' in that $Ax = \binom{L}{C \ B} \binom{u}{B} = \binom{Lu}{C}$ (using the above examples), and leads naturally to change of bases when the operator $L$ is the identity operator.

Thus to represent $u$ in the basis $C$, we just do $\binom{\mathrm{Id}}{C \ B} \binom{u}{B} = \binom{u}{C}$.

This works for operators too, if $M:U \to U$, with ordered bases $C$ and $D=(d_1,...,d_n)$ for domain and range respectively, then it follows that $\binom{M \circ L}{D \ B} = \binom{M}{D \ C} \binom{L}{C \ B}$.

So, to change the domain and range bases of $L$ to ordered bases $B',C'$ respectively, we just pre- and post-multiply $A$ by the identity operator in the appropriate bases, ie, $\binom{L}{C' \ B'} = \binom{\mathrm{Id}}{C' \ C} \binom{L}{C \ B} \binom{\mathrm{Id}}{B \ B'}$.

We note that $\binom{\mathrm{Id}}{B \ B} = I$, from which it follows that $\binom{\mathrm{Id}}{B \ B'} = \binom{\mathrm{Id}}{B' \ B}^{-1}$.

In your example, if I understand correctly, you have a basis $E$, and two other bases which I will call $\cal{B}, \cal{B}'$ respectively, and we have $B = \binom{\mathrm{Id}}{E \ {\cal B} }$, $B' = \binom{\mathrm{Id}}{E \ {\cal B}'}$, and the matrix $A$ represents some linear operator $L$ with domain and range basis $\cal{B}$, ie, $A = \binom{\mathrm{L}}{\cal{B} \ \cal{B}}$.

The goal, as I understand it, is to represent $L$ in the basis $\cal{B}'$. From the above, this would be the matrix $A' = \binom{\mathrm{L}}{\cal{B}' \ \cal{B}'}= \binom{\mathrm{Id}}{\cal{B}' \ \cal{B}} \binom{\mathrm{L}}{\cal{B} \ \cal{B}} \binom{\mathrm{Id}}{\cal{B} \ \cal{B}'}$. We note that $\binom{\mathrm{Id}}{\cal{B} \ \cal{B}'} = \binom{\mathrm{Id}}{{\cal B} \ E} \binom{\mathrm{Id}}{E \ \cal{B}'} = B^{-1} B'$, which gives $A' = (B' B^{-1}) A (B' B^{-1})^{-1}$

The row operations that you wanted to take correspond to the pre-multiplying by $B' B^{-1}$, but neglect the corresponding modifications of the domain basis.

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Since $A$ is expressible in $B$ (also $B$ is a $n\times n$ matrix), $A=BX$ for some co-efficient matrix $X$. Note that by definition $B'$ is also a invertible $n \times n$ matrix. So $A=BX=B'(B')^{-1}BX=B'X'$ where $X'=(B')^{-1}BX$ is the new co-efficient matrix in terms of the new Basis. Is this what you are looking for?

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