Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 21 September 2014

Lecture 9: Change of basis

Let $V$ be a vector space. Let $B=\{b_1,b_2,\dots \}$ be a basis of $V\text{.}$ Let $C=\{c_1,c_2,\dots \}$ be another basis of $V\text{.}$

The change of basis matrix from $B$ to $C$ is $$P=\left(p_{ij}\right)\text{given by}{c}_j=p_{1j}{b}_1+p_{2j}{b}_2+\cdots \text{.}$$ The change of basis matrix from $C$ to $B$ is $$Q=\left(q_{ij}\right)\text{given by}{b}_j=q_{1j}{c}_1+q_{2j}+\cdots \text{.}$$ Let $f:V\to V$ be a linear transformation. The matrix of $f$ with respect to $B$ is $$B_f=\left(f_{ij}^B\right)\text{given by}f\left(b_j\right)=f_{1j}^B{b}_1+f_{2j}^B{b}_2+\cdots \text{.}$$ The matrix of $f$ with respect to $C$ is $$C_f\left(f_{ij}^C\right)\text{given by}f\left(c_j\right)=f_{1j}^C{c}_1+f_{2j}^C{c}_2+\cdots \text{.}$$

(a)	$P=Q^{-1}\text{.}$
(b)	$B_f=Q{C}_f{Q}^{-1}\text{.}$

My favourite vector space $$V={ℂ}^3=\left\{\left(\begin{array}{c}{a}_1\\ a_2\\ 3_1\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{a}_1,a_2,a_3\in ℂ\}$$ has basis $$B=\{b_1,b_2,b_3\}$$ with $$b_1=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),b_2=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),b_3=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\text{.}$$ The matrix $$B_f=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$$ defines a linear transformation $f:V\to V,$ $$f\left(b_1\right)=b_2,f\left(b_2\right)=b_3,f\left(b_3\right)=b_1$$ and $$\begin{array}{ccc}{\displaystyle f\left(\begin{array}{c}3\\ 6\\ 21\end{array}\right)}& =& {\displaystyle f(3b_1+6b_2+21b_3)}\\ & =& {\displaystyle 3f\left(b_1\right)+6f\left(b_2\right)+21f\left(b_3\right)}\\ & =& {\displaystyle 3b_2+6b_3+21b_1}\\ & =& {\displaystyle \left(\begin{array}{c}21\\ 3\\ 6\end{array}\right)\text{.}}\end{array}$$ Another basis of $V$ is $$C=\{c_1,c_2,c_3\}$$ with $$c_1=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),c_2=\left(\begin{array}{c}1\\ \frac{-1+\sqrt{3}i}{2}\\ \frac{-1-\sqrt{3}i}{2}\end{array}\right),c_2=\left(\begin{array}{c}1\\ \frac{-1-\sqrt{3}i}{2}\\ \frac{-1+\sqrt{3}i}{2}\end{array}\right)\text{.}$$ Then $$\begin{array}{ccc}{\displaystyle c_1}& =& {\displaystyle b_1+b_2+b_3,}\\ {\displaystyle c_2}& =& {\displaystyle b_1+\left(\frac{-1+\sqrt{3}i}{2}\right)b_2+\left(\frac{-1-\sqrt{3}i}{2}\right)b_3,}\\ {\displaystyle c_3}& =& {\displaystyle b_1+\left(\frac{-1-\sqrt{3}i}{2}\right)b_2+\left(\frac{-1+\sqrt{3}i}{2}\right)b_3,}\end{array}$$ and $$\begin{array}{ccc}{\displaystyle b_1}& =& {\displaystyle \frac{1}{3}(c_1+c_2+c_3),}\\ {\displaystyle b_2}& =& {\displaystyle \frac{1}{3}(c_1+\frac{-1-\sqrt{3}i}{2}{c}_2+\frac{-1+\sqrt{3}i}{2}{c}_3),}\\ {\displaystyle b_3}& =& {\displaystyle \frac{1}{3}(c_1+\frac{-1+\sqrt{3}i}{2}{c}_2+\frac{-1-\sqrt{3}i}{2}{c}_3)\text{.}}\end{array}$$ Helpful: Let $\zeta =\frac{-1+\sqrt{3}i}{2}$ and note that $$\begin{array}{ccc}{\displaystyle {\zeta }^2}& =& {\displaystyle {\left(\frac{-1+\sqrt{3}i}{2}\right)}^2=\frac{1-2\sqrt{3}i-3}{4}=\frac{-1-\sqrt{3}i}{2}\text{and}}\\ {\displaystyle {\zeta }^3}& =& {\displaystyle \frac{(-1-\sqrt{3}i)}{2}\frac{(-1+\sqrt{3}i)}{2}=\frac{1+3}{4}=1,}\end{array}$$ and $$1+\zeta +{\zeta }^2=0\text{.}$$ So $$\begin{array}{ccc}{\displaystyle c_1}& =& {\displaystyle b_1+b_2+b_3,}\\ {\displaystyle c_2}& =& {\displaystyle b_1+\zeta b_2+{\zeta }^2{b}_3,}\\ {\displaystyle c_3}& =& {\displaystyle b_1+{\zeta }^2{b}_2+\zeta b_3}\end{array}$$ and $$\begin{array}{ccc}{\displaystyle \frac{1}{3}(c_1+c_2+c_3)}& =& {\displaystyle \frac{1}{3}(\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+\left(\begin{array}{c}1\\ \zeta \\ {\zeta }^2\end{array}\right)+\left(\begin{array}{c}1\\ {\zeta }^2\\ \zeta \end{array}\right))=\frac{1}{3}\left(\begin{array}{c}3\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)=b_1,}\\ {\displaystyle \frac{1}{3}(c_1+{\zeta }^2{c}_2+\zeta c_3)}& =& {\displaystyle \frac{1}{3}(\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+{\zeta }^2\left(\begin{array}{c}1\\ \zeta \\ {\zeta }^2\end{array}\right)+\zeta \left(\begin{array}{c}1\\ {\zeta }^2\\ \zeta \end{array}\right))=\frac{1}{3}(\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+\left(\begin{array}{c}{\zeta }^2\\ 1\\ \zeta \end{array}\right)+\left(\begin{array}{c}\zeta \\ 1\\ {\zeta }^2\end{array}\right))=\frac{1}{3}\left(\begin{array}{c}0\\ 3\\ 0\end{array}\right)=b_2,}\\ {\displaystyle \frac{1}{3}(c_1+\zeta c_2+{\zeta }^2{c}_3)}& =& {\displaystyle \frac{1}{3}(\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+\zeta \left(\begin{array}{c}1\\ \zeta \\ {\zeta }^2\end{array}\right)+{\zeta }^2\left(\begin{array}{c}1\\ {\zeta }^2\\ \zeta \end{array}\right))=\frac{1}{3}(\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+\left(\begin{array}{c}\zeta \\ {\zeta }^2\\ 1\end{array}\right)+\left(\begin{array}{c}{\zeta }^2\\ \zeta \\ 1\end{array}\right))=\frac{1}{3}\left(\begin{array}{c}0\\ 0\\ 3\end{array}\right)=b_{\mathrm{b}}\text{.}}\end{array}$$ The change of basis matrix from $B$ to $C$ is $$P=\left(\begin{array}{ccc}1& 1& 1\\ 1& \zeta & {\zeta }^2\\ 1& {\zeta }^2& \zeta \end{array}\right)$$ and the change of basis matrix from $C$ to $B$ is $$Q=\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}{\zeta }^2& \frac{1}{3}\zeta \\ \frac{1}{3}& \frac{1}{3}\zeta & \frac{1}{3}{\zeta }^2\end{array}\right)$$ and $$PQ=\left(\begin{array}{ccc}1& 1& 1\\ 1& \zeta & {\zeta }^2\\ 1& {\zeta }^2& \zeta \end{array}\right)\left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}{\zeta }^2& \frac{1}{3}\zeta \\ \frac{1}{3}& \frac{1}{3}\zeta & \frac{1}{3}{\zeta }^2\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\text{.}$$ So $P^{-1}=Q\text{.}$

The matrix of $f$ with respect to $C\text{:}$ $$\begin{array}{ccc}{\displaystyle f\left(c_1\right)}& =& {\displaystyle f\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=c_1,}\\ {\displaystyle f\left(c_2\right)}& =& {\displaystyle f\left(\begin{array}{c}1\\ \zeta \\ {\zeta }^2\end{array}\right)=\left(\begin{array}{c}{\zeta }^2\\ 1\\ \zeta \end{array}\right)={\zeta }^2\left(\begin{array}{c}1\\ \zeta \\ {\zeta }^2\end{array}\right)={\zeta }^2{c}_2,}\\ {\displaystyle f\left(c_3\right)}& =& {\displaystyle f\left(\begin{array}{c}1\\ {\zeta }^2\\ \zeta \end{array}\right)=\left(\begin{array}{c}\zeta \\ 1\\ {\zeta }^2\end{array}\right)=\zeta \left(\begin{array}{c}1\\ {\zeta }^2\\ \zeta \end{array}\right)=\zeta c_1\text{.}}\end{array}$$ So the matrix of $f$ with respect to $C$ is $$C_f=\left(\begin{array}{ccc}1& 0& 0\\ 0& {\zeta }^2& 0\\ 0& 0& \zeta \end{array}\right)\text{.}$$ Magic: $$\begin{array}{ccc}{\displaystyle Q{B}_{f}P}& =& {\displaystyle \left(\begin{array}{ccc}\frac{1}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}{\zeta }^2& \frac{1}{3}\zeta \\ \frac{1}{3}& \frac{1}{3}\zeta & \frac{1}{3}{\zeta }^2\end{array}\right)\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{ccc}1& 1& 1\\ 1& \zeta & {\zeta }^2\\ 1& {\zeta }^2& \zeta \end{array}\right)}\\ & =& {\displaystyle \frac{1}{3}\left(\begin{array}{ccc}1& 1& 1\\ 1& {\zeta }^2& \zeta \\ 1& \zeta & {\zeta }^2\end{array}\right)\left(\begin{array}{ccc}1& {\zeta }^2& \zeta \\ 1& 1& 1\\ 1& \zeta & {\zeta }^2\end{array}\right)}\\ & =& {\displaystyle \frac{1}{3}\left(\begin{array}{ccc}1& 0& 0\\ 0& 3{\zeta }^2& 0\\ 0& 0& 3\zeta \end{array}\right)}\\ & =& {\displaystyle \left(\begin{array}{ccc}1& 0& 0\\ 0& {\zeta }^2& 0\\ 0& 0& \zeta \end{array}\right)}\\ & =& {\displaystyle C_f\text{.}}\end{array}$$

Let $f:V\to V$ be a linear transformation. Let $B$ and $C$ be bases of $V\text{.}$ Let $P$ be the change of basis matrix from $B$ to $C,$
$Q$ the change of basis matrix from $C$ to $B,$
$B_f$ the matrix of $f$ with respect to $B,$
$C_f$ the matrix of $f$ with respect to $C\text{.}$ Then

(a)	$P=Q^{-1},$
(b)	$B_f=Q{C}_f{Q}^{-1}\text{.}$

		Proof.
	(a) To show: (aa) $PQ=1,$ (ab) $QP=1\text{.}$ (aa) We know: $$\begin{array}{ccc}{\displaystyle P}& =& {\displaystyle \left(p_{ij}\right)\text{with}{c}_j=p_{1j}{b}_1+p_{2j}{b}_2+\cdots ,}\\ {\displaystyle Q}& =& {\displaystyle \left(q_{k\ell }\right)\text{with}{b}_{\ell }=q_{1\ell }{c}_1+q_{2\ell }{c}_2+\cdots \text{.}}\end{array}$$ To show: $PQ=1\text{.}$ To show: (aaa) ${\left(PQ\right)}_{ii}=1,$ (aab) ${\left(PQ\right)}_{ij}=0$ if $i\ne j\text{.}$ (aaa) $${\left(PQ\right)}_{ii}=p_{i1}{q}_{1i}+p_{i2}{q}_{2i}+p_{i3}{q}_{3i}+\cdots$$ since $$\begin{array}{ccc}{\displaystyle b_i}& =& {\displaystyle q_{1i}{c}_1+q_{2i}{c}_2+q_{3i}{c}_3+\cdots }\\ & =& {\displaystyle q_{1i}(p_{11}{b}_1+p_{21}{b}_2+p_{31}{b}_3+\cdots )}\\ & & {\displaystyle +q_{2i}(p_{12}{b}_1+p_{22}{b}_2+p_{32}{b}_3+\cdots )}\\ & & {\displaystyle +q_{3i}(p_{13}{b}_1+p_{23}{b}_2+p_{33}{b}_3+\cdots )}\\ & & {\displaystyle ⋮}\\ & =& {\displaystyle (q_{1i}{p}_{11}+q_{2i}{p}_{12}+q_{3i}{p}_{13}+\cdots )b_1+}\\ & & {\displaystyle (q_{1i}{p}_{21}+q_{2i}{p}_{22}+q_{3i}{p}_{23}+\cdots )b_2+\cdots \text{.}}\end{array}$$ If we use sum notation $$\begin{array}{ccc}{\displaystyle b_j}& =& {\displaystyle \sum _{\ell }{q}_{\ell j}{c}_{\ell }}\\ & =& {\displaystyle \sum _{\ell }{q}_{\ell j}\left(\sum _m{p}_{m\ell }{b}_m\right)}\\ & =& {\displaystyle \sum _{\ell ,m}{p}_{m\ell }{q}_{\ell j}{b}_m\text{.}}\end{array}$$ So $$\sum _{\ell }{p}_{m\ell }{q}_{\ell j}=0$$ if $m\ne j$ and $$\sum _{\ell }{p}_{j\ell }{q}_{\ell j}=1\text{.}$$ $\square$

Notes and References

These are a typed copy of Lecture 9 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 12, 2011.

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