# Linear Algebra: Change of Basis

Let $A[a,b,c]$ and $B[d,e,f]$ be two non-standard bases. I have to find the $3\times3$ matrix that will convert a vector defined in terms of $A$ to $B$.

My solution is:

Let's assume a standard basis $O[e_1,e_2,e_3]$.

Now, the rotation matrix $R(A\to O)$ to change basis from $A$ to $O$ is:

$\begin{bmatrix} ax & bx & cx \\ ay & by & cy \\ az & bz & cz \end{bmatrix}$

and $R(O\to B)$ is

$\begin{bmatrix} ax & ay & az \\ bx & by & bz \\ cx & cy & cz \end{bmatrix}.$

So, $R(A \to B)=R(A \to O)\cdot R(O\to B)$.

Is this the correct solution?

Thanks a lot.

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I see no reason you should expect a rotation matrix. Two arbitrary bases are just related by multiplication by an invertible matrix (of which many are not rotations!)

Moreover, when thinking about this sort of question for the first time it's wise to develop some notation which denotes the coordinate vectors for differing choices of basis. I'm not seeing this in your post.

A typical notation goes like this: if $v \in \mathbb{R}^3$ and $v = c_1f_1+c_2f_2+c_3f_3$ then $\Phi_{\beta}(v)=[v]_{\beta} = [c_1,c_2,c_3]^T$ where $\beta = \{ f_1,f_2,f_3 \}$ is a (possibly) nonstandard basis. You can derive all sorts of short-cut formulas for $\mathbb{R}^3$ since the coordinate map $\Phi_{\beta}$ is a linear transformation on $\mathbb{R}^3$.

If you search posts about change of basis (many of which are linked over $\rightarrow$

then you'll get a better idea about how to communicate these sort of calculations. Unfortunately, at the present, I can't quite get what you're saying in the post.

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Let $\alpha=\{a,b,c\}$ and $\beta=\{d,e,f\}$ be bases for $\mathbb{R}^3$ (thinking of elements of $\mathbb{R}^3$ as column vectors). Then the matrix $A=[a|b|c]$ transitions from $\alpha$-coordinates to standard-coordinates. Likewise the basis $B=[d|e|f]$ transitions from $\beta$-coordinates to the standard-coordinates. So $B^{-1}$ goes from standard-coordinates to $\beta$-coordinates. Thus $B^{-1}A$ goes from $\alpha$-coordinates (to standard and then standard) to $\beta$-coordinates and is the matrix you're looking for.

Example: Let $\alpha=\left\{\begin{bmatrix} 1\\0\\0 \end{bmatrix},\begin{bmatrix} 1\\1\\0 \end{bmatrix},\begin{bmatrix} 1\\1\\1 \end{bmatrix}\right\}$ and $\beta=\left\{\begin{bmatrix} 1\\-1\\0 \end{bmatrix},\begin{bmatrix} 0\\1\\-1 \end{bmatrix},\begin{bmatrix} 1\\0\\1 \end{bmatrix}\right\}$.

Then ${\bf v}=\begin{bmatrix} 1\\2\\3 \end{bmatrix} = (-1)\begin{bmatrix} 1\\0\\0 \end{bmatrix}+(-1)\begin{bmatrix} 1\\1\\0 \end{bmatrix}+3\begin{bmatrix} 1\\1\\1 \end{bmatrix}$. So ${\bf v}$ in $\alpha$ coordinates is $[{\bf v}]_\alpha = \begin{bmatrix} -1\\-1\\3\end{bmatrix}$.

Let $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$. Notice that $A[{\bf v}]_\alpha = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} -1\\-1\\3\end{bmatrix} = \begin{bmatrix} 1\\2\\3\end{bmatrix}$.

Let $B=\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{bmatrix}$. Then $B^{-1} = \frac{1}{2}\begin{bmatrix} 1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix}$ and so $B^{-1}A = \begin{bmatrix} \frac{1}{2} & 0 & -\frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{3}{2} \end{bmatrix}$ goes from $\alpha$ to $\beta$-coordinates.

An alternate way to compute this change of basis matrix is to create the augmented matrix: $[B|A] = [d|e|f|a|b|c]$ and row-reduce. This will compute $[I|(B^{-1}A)]$ where $B^{-1}A$ is the desired matrix.

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