Let V be a finite dimensional vector space over field F and X, Y linear transformations from V to 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 $Y = ZXZ^{-1}$?

  • $\begingroup$ Can you retype it using LaTex? $\endgroup$ – Falang Sep 29 '13 at 5:25
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    $\begingroup$ Let $Z$ be the matrix that sends one basis to the other. $\endgroup$ – Prahlad Vaidyanathan Sep 29 '13 at 5:35

I take it that this is an exercise to help you understand the relation between linear transformations, matrices, and bases. Because of that, I'll try to be explicit these things.

So we have linear transformations $X, Y \colon V \to V$. First, look at what $[X]_{A,A}$ (and similarly $[Y]_{B,B}$ means. This is, I presume, the matrix representing the linear transformation $X$ on the basis $A$ (for both the domain and codomain).

Alright, what does that mean? Choosing a basis $A = (A_1, \dots, A_n)$ of $V$ can be seen as giving an explicit isomorphism $e_A \colon F^n \to V, (x_1, \dots, x_n) \mapsto x_1 A_1 + \dots + x_n A_n$. Furthermore, an $n \times n$ matrix (such as $[X]_{A,A}$) can be seen as an automorphism of $F^n$, given by left-multiplication with the matrix, let's call this automorpism $\mu_M \colon F^n \to F^n, (x_1, \dots, x_n) \mapsto M (x_1, \dots, x_n)$. (By the way, $(x_1, \dots, x_n)$ is a column vector; writing it like this saves space.)

Now, "$[X]_{A,A}$ is the matrix representing the linear transformation $X$ on the basis $A$" means that we have a commutative diagram $$\begin{array}{} \;\;\;\; F^{n} & \stackrel{\mu_{[X]_{A,A}}} {\longrightarrow} & \;\; F^n \\ {e_A}\downarrow \cong& & \cong \downarrow{e_A} \\ \;\;\;\;V & \stackrel{X}{\longrightarrow} & \;\;V, \end{array} $$ that is $X = e_A \circ \mu_{X_[A,A]} \circ e_A^{-1}$.

Similarly for $Y$: $$\begin{array}{} \;\;\;\; F^{n} & \stackrel{\mu_{[Y]_{B,B}}} {\longrightarrow} & \;\; F^n \\ {e_B}\downarrow \cong& & \cong \downarrow{e_B} \\ \;\;\;\;V & \stackrel{Y}{\longrightarrow} & \;\;V. \end{array} $$

Statement. There exists ordered bases $A$ and $B$ of $V$ such that $[X]_{A,A} = [Y]_{B,B}$ if and only if there is a linear automorphism $Z \colon V \to V$ with $Y = Z \circ X \circ Z^{-1}$.

Proof. $(\Rightarrow)$ Let's start by assuming that there are ordered bases $A$ and $B$ of $V$ with $[X]_{A,A} = [Y]_{B,B}$. Writing just $\mu$ for $\mu_{[X]_{A,A}} = \mu_{[Y]_{B,V}}$, we can patch the above two commutative diagrams together to give the following commutative diagram: $$\begin{array}{} \;\;\;\;V & \stackrel{X}{\longrightarrow} & \;\;V \\ {e_A}\uparrow \cong& & \cong \uparrow{e_A} \\ \;\;\;\; F^{n} & \stackrel{\mu} {\longrightarrow} & \;\; F^n \\ {e_B}\downarrow \cong& & \cong \downarrow{e_B} \\ \;\;\;\;V & \stackrel{Y}{\longrightarrow} & \;\;V. \end{array} $$ So, $Y = e_B \circ e_A^{-1} \circ X \circ e_A \circ e_B^{-1}$. Letting $Z$ be $e_B \circ e_A^{-1}$, we get $Y = Z \circ X \circ Z^{-1}$.

$(\Leftarrow)$ Conversely, let's assume an automorphism $Z \colon V \to V$ with $Y = Z \circ X \circ Z^{-1}$ exists. We can visualize this in a commutative diagram as follows: $$\begin{array}{} \;\;\; V & \stackrel{X} {\longrightarrow} & \;\; V \\ {Z}\downarrow \cong& & \cong \downarrow{Z} \\ \;\;\;V & \stackrel{Y}{\longrightarrow} & \;\;V. \end{array} $$

Now pick any basis $A$ of $V$ and we'll try to find a basis $B$ of $V$ such that $[X]_{A,A} = [Y]_{B,B}$. Patching the very first and previous diagram together, we get $$\begin{array}{} \;\;\;\; F^{n} & \stackrel{\mu_{[X]_{A,A}}} {\longrightarrow} & \;\; F^n \\ {e_A}\downarrow \cong& & \cong \downarrow{e_A} \\ \;\;\;\;V & \stackrel{X}{\longrightarrow} & \;\;V \\ {Z}\downarrow \cong& & \cong \downarrow{Z} \\ \;\;\;\;V & \stackrel{Y}{\longrightarrow} & \;\;V. \end{array} $$ We now pick the basis $B$ in such a way that the composition $Z \circ e_A$ is exactly $e_B$. Therefore, we take $B_1 = Z(A_1), \dots, B_n = Z(A_n)$. This means that the outer square in the previous diagram really is $$\begin{array}{} \;\;\;\; F^{n} & \stackrel{\mu_{[X]_{A,A}}} {\longrightarrow} & \;\; F^n \\ {e_B}\downarrow \cong& & \cong \downarrow{e_B} \\ \;\;\;\;V & \stackrel{Y}{\longrightarrow} & \;\;V. \end{array} $$ The top arrow, however, is also $\mu_{[Y]_{B,B}}$ because this is exactly the diagram that defines what "$[Y]_{B,B}$ is the matrix representing $Y$ on basis $B$" means. Therefore $\mu_{[X]_{A,A}} = \mu_{[Y]_{B,B}}$ and therefore also $[X]_{A,A} = [Y]_{B,B}$.


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