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 13: Block decomposition

Let $V$ be a vector space. Let $B=\{b_1,\dots ,b_k\}$ and $C=\{c_1,\dots ,c_{\ell }\}$ be bases of $V\text{.}$ Then $k=\ell \text{.}$

		Proof.
	Proof by contradiction. Assume $k<\ell \text{.}$ Let $C\prime =\{c_2,\dots ,c_{\ell }\}\text{.}$ Then there exists $b_j\in B$ such that $b_j\notin \text{span}\left(C\prime \right)\text{.}$ Proof. If all $b_1,\dots ,b_k\in \text{span}\left(C\prime \right)$ then $$\text{span}\left(B\right)\subseteq \text{span}\left(C\prime \right)\text{and}V\subseteq \text{span}\left(C\prime \right)\text{.}$$ Since $c_1\in V$ then $c_1\in \text{span}\left(C\prime \right)\text{.}$ So $c_1={\gamma }_2{c}_2+\cdots +{\gamma }_{\ell }{c}_{\ell }$ for some ${\gamma }_2,\dots ,{\gamma }_{\ell }\in 𝔽\text{.}$ So $0=-c_1+{\gamma }_2{c}_2+\cdots +{\gamma }_{\ell }{c}_{\ell }\text{.}$ This is a contradiction to the linear independence of $C\text{.}$ So there exists $b_j$ such that $b_j\notin \text{span}\left(C\prime \right)\text{.}$ $\square$ Reindex $B$ so that $b_1\notin \text{span}\left(C\prime \right)$ and let $C_1=\{b_1,c_2,c_3,\dots ,c_{\ell }\}\text{.}$ Claim $C_1$ is a basis of $V\text{.}$ Proof of claim. To show: (a) $C_1$ is linearly independent. (b) $\text{span}\left(C_1\right)=V\text{.}$ (a) To show: If ${\alpha }_1{b}_1+{\alpha }_2{c}_2+\cdots +{\alpha }_{\ell }{c}_{\ell }=0$ then ${\alpha }_1=0,$ ${\alpha }_2=0,$ $\dots ,$ ${\alpha }_{\ell }=0\text{.}$ Assume ${\alpha }_1{b}_1+{\alpha }_2{c}_2+\cdots +{\alpha }_{\ell }{c}_{\ell }=0\text{.}$ Case 1 $\alpha \ne 0\text{.}$ Then $b_1=-\frac{{\alpha }_2}{{\alpha }_1}{c}_2-\frac{{\alpha }_3}{{\alpha }_1}{c}_3-\cdots -\frac{{\alpha }_{\ell }}{{\alpha }_1}{a}_{\ell }\text{.}$ So $b_1\in \text{span}\left(C\prime \right),$ which is a contradiction to the choice of $b_1\text{.}$ So ${\alpha }_1=0\text{.}$ Case 2 ${\alpha }_1=0\text{.}$ Then ${\alpha }_2{c}_2+\cdots +{\alpha }_{\ell }{c}_{\ell }=0\text{.}$ Since $C$ is linearly independent $${\alpha }_2=0,{\alpha }_3=0,\dots ,{\alpha }_{\ell }=0\text{.}$$ (b) To show: $V=\text{span}\{b_1,c_2,\dots ,c_{\ell }\}\text{.}$ To show: $V\subseteq \text{span}\{b_1,c_2,\dots ,c_{\ell }\}\text{.}$ To show: $\text{span}\{c_1,c_2,\dots ,c_{\ell }\}\subseteq \text{span}\{b_1,c_2,\dots ,c_{\ell }\}\text{.}$ To show: $c_1\in \text{span}\{b_1,c_2,\dots ,c_{\ell }\}\text{.}$ Since $C$ is a basis there exist ${\alpha }_1,\dots ,{\alpha }_{\ell }\in 𝔽$ with $b_1={\alpha }_1{c}_1+{\alpha }_2{c}_2+\cdots +{\alpha }_{\ell }{c}_{\ell }\text{.}$ Case 1 ${\alpha }_1=0\text{.}$ Then $b_1={\alpha }_2{c}_2+\cdots +{\alpha }_{\ell }{c}_{\ell }$ and $b_1\in \text{span}\left(C\prime \right)\text{.}$ This is a contradiction to the choice of $b_1\text{.}$ So ${\alpha }_1\ne 0\text{.}$ Case 2 ${\alpha }_2\ne 0\text{.}$ Then $c_1={\alpha }_1^{-1}(b_1-{\alpha }_2{c}_2-\cdots -{\alpha }_{\ell }{c}_{\ell })\text{.}$ So $c_1\in \text{span}\{b_1,c_2,\dots ,c_{\ell }\}\text{.}$ So $C_1=\{b_1,c_2,\dots ,c_{\ell }\}$ is a basis of $V\text{.}$ $\square$ Let $C_1^{\prime }=\{b_1,c_3,\dots ,c_{\ell }\}$ and let $b_j\in B$ such that $b_j\notin \text{span}\left(C_1^{\prime }\right)\text{.}$ Reindex $B$ so that $b_2\notin \text{span}\left(C_1^{\prime }\right)\text{.}$ Let $C_2=\{b_1,b_2,c_3,\dots ,c_{\ell }\}\text{.}$ Then, by a proof as for $C_1$ above, $C_2$ is a basis of $V\text{.}$ Continue this process to obtain $$C_k=\{b_1,b_2,\dots ,b_k,c_{k+1},c_{k+2},\dots ,c_{\ell }\}$$ which is a basis of $V\text{.}$ Since $B$ is a basis of $V,$ then $$c_{k+1}={\alpha }_1{b}_1+{\alpha }_2{b}_2+\cdots +{\alpha }_k{b}_k,$$ for some ${\alpha }_1,\dots ,{\alpha }_k\in 𝔽\text{.}$ So $0={\alpha }_1{b}_1+\cdots +{\alpha }_k{b}_k-c_{k+1}\text{.}$ This contradicts the linear independence of $C_k\text{.}$ So $k\nless \ell \text{.}$ A similar argument shows $\ell \nless k\text{.}$ So $k=\ell \text{.}$ $\square$

Let $V$ be a vector space and let $f:V\to V$ be a linear transformation. Let $m\left(t\right)$ be the minimal polynomial of $f\text{.}$ Assume that $$m\left(t\right)=p\left(t\right)q\left(t\right)$$ with $\text{gcd}(p,q)=1\text{.}$ Let $k\left(t\right)$ and $\ell \left(t\right)$ be such that $$1=p\left(t\right)k\left(t\right)+q\left(t\right)\ell \left(t\right)\text{.}$$ Let $$U=p\left(f\right)k\left(f\right)V\text{and}W=q\left(f\right)\ell \left(f\right)V\text{.}$$ Then $$V=U\oplus W\text{.}$$

Let $$A=\left(\begin{array}{cc}2& 0\\ 1& 2\\ & & 3& 0\\ & & 1& 3\end{array}\right)\text{.}$$ The minimal polynomial of $A$ is $$m\left(t\right)={(t-2)}^2{(t-3)}^2$$ and $p\left(t\right)={(t-2)}^2,$ $q\left(t\right)={(t-3)}^2\text{.}$ $$p\left(t\right)=t^2-4t+4\text{and}q\left(t\right)=t^2-6t+9\text{.}$$ $$\begin{array}{ccc}{\displaystyle t^2-4t+4}& =& {\displaystyle (t^2-6t+9)+(2t-5)}\\ {\displaystyle t^2-6t+9}& =& {\displaystyle (2t-5)(\frac{1}{2}t-\frac{7}{4})+\frac{1}{4}}\end{array}$$ since $$t^2-4t+4=(t^2-6t+9)+(2t-5)\text{.}$$ $$\begin{array}{cc}& \frac{1}{2}t-\frac{7}{4}\\ 2t-5& \overline{|\hspace{0.17em}{t}^2-6t+9\phantom{}}\\ & \phantom{|\hspace{0.17em}}{t}^2-\frac{5}{2}t\\ & \overline{\phantom{|\hspace{0.17em}{t}^2}-\frac{7}{2}t+9\phantom{}}\\ & \phantom{|\hspace{0.17em}{t}^2}-\frac{7}{2}t+\frac{35}{4}\phantom{0.5}\\ & \phantom{|\hspace{0.17em}{t}^2}\overline{\phantom{-\frac{7}{2}t+}\frac{1}{4}\phantom{}}\end{array}$$ So $$t^2-6t+9=(\frac{1}{2}t-\frac{7}{4})(2t-5)+\frac{1}{4}\text{.}$$ So $$\begin{array}{ccc}{\displaystyle 1}& =& {\displaystyle 4(t^2-6t+9)-(2t-7)(2t-5)}\\ & =& {\displaystyle 4(t^2-6t+9)-(2t-7)((t^2-4t+4)-(t^2-6t+9))}\\ & =& {\displaystyle (2t-3)(t^2-6t+9)-(2t-7)(t^2-4t+4)}\\ & =& {\displaystyle (2t-3){(t-3)}^2-(2t-7){(t-2)}^2\text{.}}\end{array}$$ Then $$\begin{array}{ccc}{\displaystyle (2A-3){(A-3)}^2}& =& {\displaystyle \left(\begin{array}{cc}1& 0\\ 2& 1\\ & & 3& 0\\ & & 2& 3\end{array}\right){\left(\begin{array}{c}-1\\ 1& -1\\ & & 0\\ & & 1& 0\end{array}\right)}^2}\\ & =& {\displaystyle \left(\begin{array}{c}1\\ 2& 1\\ & & 3\\ & & 2& 3\end{array}\right)\left(\begin{array}{cc}1& 0\\ -2& 1\\ & & 0& 0\\ & & 0& 0\end{array}\right)}\\ & =& {\displaystyle \left(\begin{array}{cc}1& 0\\ 0& 1\\ & & 0& 0\\ & & 0& 0\end{array}\right)}\end{array}$$ and $$\begin{array}{ccc}{\displaystyle -(2A-7){(A-2)}^2}& =& {\displaystyle \left(\begin{array}{cc}3& 0\\ -2& 3\\ & & 1& 0\\ & & -2& 1\end{array}\right){\left(\begin{array}{cc}0& 0\\ 1& 0\\ & & 1& 0\\ & & 1& 1\end{array}\right)}^2}\\ & =& {\displaystyle \left(\begin{array}{cc}3& 0\\ -2& 3\\ & & 1& 0\\ & & -2& 1\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 0\\ & & 1& 0\\ & & 2& 1\end{array}\right)}\\ & =& {\displaystyle \left(\begin{array}{cc}0& 0\\ 0& 0\\ & & 1& 0\\ & & 0& 1\end{array}\right)\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 13 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 23, 2011.

page history