Let $V$ be a finite dimensional vector space over field $\mathbb{F}$ and $X, Y$ linear transformations from $V \mapsto V$. When do there exist ordered bases $A$ and $B$ for $V$ such that $[X]_{A,A}$ = $[Y]_{B,B}$? Prove such bases exist if and only if there is an invertible linear transformation $Z : V \rightarrow V $ such that $Y = ZXZ^{-1}$.

I tried it in one direction first, proving you get $Y = ZXZ^{-1}$ if you assume that equation is true. I'm not sure how to exactly manipulate the abstract bases. Would you let a transformation map from A to B and compose functions together to get $Y = ZXZ^{-1}$?

  • $\begingroup$ do you know about change of basis matrices? $\endgroup$ – ILikeMath Sep 29 '13 at 16:52
  • $\begingroup$ Yes would I have to somehow operate it with X and Y? $\endgroup$ – Shawn Sep 29 '13 at 17:29

Let $n \in \mathbb{N}$ be the dimension of $V$ and $Rep_A, Rep_B:V \to \mathbb{F}^n$ be maps where if $\vec{v} \in V$, $A=\langle \vec{\alpha}_1, \cdots, \vec{\alpha}_n\rangle$ and $B=\langle \vec{\beta}_1, \cdots, \vec{\beta}_n\rangle$ then

$$Rep_B{}(\vec{v})=\left( {\begin{array}{*{20}{c}} {{c_1}} \\ \vdots \\ {{c_n}} \\ \end{array}} \right)$$

and $\vec{v}=c_1\vec{\beta}_1+\cdots +c_n\vec{\beta}_n$. The same with $A$ (you can show that $Rep_A, Rep_B$ are linear and nosingular then they have inverse). Now if $[X]_{A,A}=[Y]_{B,B}$ then


for all $i \in \{ 1, \cdots, n \}$, so


for all $i \in \{ 1, \cdots, n \}$. We define $Z:V\to V$ as $Z(\vec{\alpha}_i)=\vec{\beta}_i$, then $Z^{-1}(\vec{\beta}_i)=\vec{\alpha}_i$ for all $i \in \{ 1,\cdots, n\}$ (you have to show that $Z$ is linear and nonsingular).

Now we need to show that $Y=Z\circ X\circ Z^{-1}$ equivalent to $Y(\vec{v})=(Z\circ X\circ Z^{-1})(\vec{v})$ for all $\vec{v} \in V$. Let $\vec{v} \in V$ any vector, then there are $c_1,\cdots,c_n \in \mathbb{F}$ such that $\vec{v}=c_1\vec{\beta}_1+\cdots +c_n\vec{\beta}_n$, so

$$\begin{align}Z(X(Z^{-1}(\vec{v})))&=Z(X(Z^{-1}(c_1\vec{\beta}_1+\cdots +c_n\vec{\beta}_n)))\\&=c_1Z(X(Z^{-1}(\vec{\beta}_1))) + \cdots + c_nZ(X(Z^{-1}(\vec{\beta}_n)))\end{align} \space \space (*)$$

because $Z^{-1},X,Z$ are linear. We fix $i\in \{ 1, \cdots,n\}$, then


but there are $d_1,\cdots,d_n \in \mathbb{F}$ such that $Y(\vec{\beta}_i)=d_1\vec{\beta}_1+\cdots +d_n\vec{\beta}_n$ because $Y(\vec{\beta}_i)\in V$, then

$$Rep_B(Y(\vec{\beta}_i))=\left( {\begin{array}{*{20}{c}} {{d_1}} \\ \vdots \\ {{d_n}} \\ \end{array}} \right)$$


$$\begin{align}c_iZ(X(Z^{-1}(\vec{\beta}_i)))&=c_iZ(Rep_A^{-1}(\left( {\begin{array}{*{20}{c}} {{d_1}} \\ \vdots \\ {{d_n}} \\ \end{array}} \right)))\\&=c_iZ(d_1\vec{\alpha}_1+\cdots + d_n\vec{\alpha}_n)\\&=c_i\left(d_1Z(\vec{\alpha}_1)+\cdots+d_nZ(\vec{\alpha}_n)\right)\\ &=c_i\left(d_1\vec{\beta}_1+\cdots+d_n\vec{\beta}_n\right)\\&=c_iY(\vec{\beta}_i)\end{align}$$

because $Z$ is linear. Therefore from (*)

$$\begin{align}Z(X(Z^{-1}(\vec{v})))&=c_1Z(X(Z^{-1}(\vec{\beta}_1))) + \cdots + c_nZ(X(Z^{-1}(\vec{\beta}_n)))\\ &=c_1Y(\vec{\beta}_1) + \cdots + c_nY(\vec{\beta}_n)\\ &=Y(c_1\vec{\beta}_1 + \cdots + c_n\vec{\beta}_n)\\ &=Y(\vec{v}),\end{align}$$

then $Y=Z \circ X \circ Z^{-1}$.


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