PS01--rep-theory--solutions.md

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title: |
 ProblemSet 1 -- Linear algebra and representations **Solutions**
author: George McNinch
date: 2024-01-29

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$F$ denotes an algebraically closed field of characteristic 0. If you like, you can suppose that $F = \mathbf{C}$ is the field of complex numbers.

1. Let $V$ be a finite dimensional vector space over the field $F$. Suppose that $\phi,\psi:V \to V$ are linear maps. Let $\lambda \in F$ be an eigenvalue of $\phi$ and write $W$ for the $\lambda$-eigenspace of $\phi$; i.e. $$W = \{v \in V \mid \phi(v) = \lambda v \}.$$ If $\phi \psi = \psi \phi$ show that $W$ is invariant under $\psi$ -- i.e. show that $\psi(W) \subseteq W$.

   ::: {.solution} Solution: Let $w \in W$. We must show that $x=\psi(w) \in W$. To do this, we must establish that $x=\psi(w)$ is a $\lambda$-eigenvector for $\phi$.

   We have $$\begin{align*} \phi(x) &= \phi(\psi(x)) & \\ &= \psi(\phi(w)) & \text{since $\phi \circ \psi = \psi \circ \phi$} \\ &= \psi(\lambda w) & \text{since $w$ is a $\lambda$-eigenvector} \\ &= \lambda \psi(w) & \text{since $\psi$ is linear} \\ &= \lambda x \end{align*}$$ This completes the proof. :::

2. Let $n \in \mathbf{N}$ be a non-zero natural number, and let $V$ be an $n$ dimensional $F$-vector space with a given basis $e_1,e_2,\cdots,e_n$.

   Consider the linear transformation $T:V \to V$ given by the rule $$Te_i = e_{i+1 \pmod n}.$$ In other words $$Te_i = \left \{ \begin{matrix} e_{i+1} & i < n \\ e_1 & i =n \end{matrix} \right ..$$

   a. Show that $T$ is invertible and that $T^n = \operatorname{id}_V$.

   ::: {.solution} To check that $T^n = \operatorname{id}_V$, we check that $T^n(e_i) = e_i$ for $1 \le i \le n$.

   From the definition, it follows by induction on the natural number $m$ that $$T^m(e_i) = e_{i+m \pmod n}.$$ Thus $T^n(e_i) = e_{i+n\pmod n} = e_i$. Since this holds for every $i$, conclude $T^n = \operatorname{id}_V$.

   Now $T$ is invertible since its inverse is given by $T^{n-1}$. :::

   b. Consider the vector $v_0 = \displaystyle \sum_{i=1}^n e_i$. Show that $v_0$ is a $1$-eigenvector for $T$.

   ::: {.solution} We compute $$\begin{align*} T(v_0) &= T\left(\sum_{i=1}^n e_i\right) = \sum_{i=1}^n T(e_i) \\ &= \sum_{i=1}^n e_{i+1 \pmod n} \\ &= \sum_{j=2}^{n+1} e_{j \pmod n} & \text{(let $j = i+1$)} \\ &= \sum_{j=1}^{n} e_{j \pmod n} = v_0 \\ \end{align*}$$ Thus $T(v_0) = v_0$ so indeed $v_0$ is a $1$-eigenvector. :::

   Let $\zeta \in F$ be a primitive $n$-th root of unity. (e.g. if you assume $F = \mathbf{C}$, you may as well take $\zeta = e^{2\pi i/n}$).

   c. Let $v_1 = \displaystyle \sum_{i=1}^n \zeta^i e_i$. Show that $v_1$ is a $\zeta^{-1}$-eigenvector for $T$.

   ::: {.solution} We compute $$\begin{align*} T(v_1) &= T\left(\sum_{i=1}^n \zeta^i e_i\right) \\ & = \sum_{i=1}^n \zeta^iT(e_i) \\ &= \sum_{i=1}^n \zeta^i e_{i+1 \pmod n} \\ &= \sum_{j=2}^{n+1} \zeta^{j-1} e_{j \pmod n} & \text{(let $j = i+1$)} \\ &= \zeta^{-1} \sum_{j=2}^{n+1} \zeta^{j} e_{j \pmod n} \\ &= \zeta^{-1} \sum_{j=1}^{n} \zeta^j e_{j \pmod n} & \text{(since $\zeta^j = \zeta^{j \pmod n}$ $\forall j$)}\\ & = \zeta^{-1} v_1 \\ \end{align*}$$ Thus $T(v_1) = \zeta^{-1} v_1$ so indeed $v_0$ is a $\zeta^{-1}$-eigenvector. :::

   d. More generally, let $0 \le j < n$ and let $$v_j = \sum_{i=1}^n \zeta^{ij} e_i.$$ Show that $v_j$ is a $\zeta^{-j}$-eigenvector for $T$.

   ::: {.solution} The calcuation in the solution to part (c) is valid for any $n$-th root of unity unity $\zeta$. Applying this calculation for $\zeta^j$ shows that $v_j$ is a $\zeta^{-j}$-eigenvector for $T$ as required. :::

   e. Conclude that $v_0,v_1,\cdots,v_{n-1}$ is a basis of $V$ consisting of eigenvectors for $T$, so that $T$ is diagonalizable.

   Hint: You need to use the fact that eigenvectors for distinct eigenvalues are linearly independent.

   What is the matrix of $T$ in this basis?

   ::: {.solution}

   Since eigenvectors for distinct eigenvalues are linearly independent, conclude that the vectors $\mathcal{B}=\{v_0,v_1,\cdots,v_{n-1}\}$ are linearly independent. Since there $n$ vectors in $\mathcal{B}$ and since $\dim V = n$, conclude that $\mathcal{B}$ is a basis for $V$.

   The matrix of $T$ in the basis $\mathcal{B}$ is given by $$[T]_{\mathcal{B}} = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & \zeta^{-1} & 0 & \cdots & 0 \\ 0 & 0 & \zeta^{-2} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \zeta^{-n+1} \end{bmatrix}$$

   (This form explains why an $n\times n$ matrix $M$ is diagonalizable iff $F^n$ has a basis of eigenvectors for $M$). :::

3. Let $G = \mathbb{Z}/3\mathbb{Z}$ be the additive group of order $3$, and let $\zeta$ be a primitive $3$rd root of unity in $F$.

   To define a representation $\rho:G \to \operatorname{GL}_n(F)$, it is enough to find a matrix $M \in \operatorname{GL}_n(F)$ with $M^3 = 1$; in turn, $M$ determines a representation $\rho$ by the rule $\rho(i + 3\mathbb{Z}) = M^i$.

   Consider the representation $\rho_1 : G \to \operatorname{GL}_3(F)$ given by the matrix $$\rho_1(1 + 3\mathbb{Z}) = M_1 = \begin{bmatrix} 1 & 0 & 0\\ 0 & \zeta & 0 \\ 0 & 0 & \zeta^2 \end{bmatrix}$$ and consider the representation $\rho_2:G \to \operatorname{GL}_3(F)$ given by the matrix $$\rho_2(1 + 3\mathbb{Z}) = M_2 = \begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}.$$

   Show that the representations $\rho_1$ and $\rho_2$ are equivalent (alternative terminology: are isomorphic). In other words, find a linear bijection $\Phi:F^3 \to F^3$ with the property that $$\Phi(\rho_2(g)v) = \rho_1(g)\Phi(v)$$ for every $g \in G$ and $v \in F^3$.

   Hint: First find a basis of $F^3$ consisting of eigenvectors for the matrix $M_2$.

   ::: {.solution} The matrix $M_1$ is diagonal, which is to say that the standard basis vectors $e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ are eigenvectors for $M_1$ with respective eigenvalues $1,\zeta,\zeta^2$.

   By the work in problem 2, we see that $$v_1 = e_1 + e_2 + e_3, \quad v_2 = e_1 + \zeta e_2 + \zeta^2 e_3, \quad v_3 = e_1 + \zeta^2 e_2 + \zeta e_3$$ are eigenvectors for $M_2$ with respective eigenvalues $1,\zeta^2,\zeta$.

   Now let $\Phi:F^3 \to F^3$ be the linear transformation for which $$\Phi(e_1) = v_1, \quad \Phi(e_2) =v_3, \quad \Phi(e_3) = v_2$$.

   We claim that $\Phi$ defines an isomorphism of $G$-representations $$(\rho_1,F^3) \xrightarrow{\sim} (\rho_2,F^3).$$

   We must check that $\Phi(\rho_1(g)v) = \rho_2(g)\Phi(v)$ for all $g \in G$ and all $v \in F^3$.

   Since $G$ is cyclic it suffices to check that $$(\clubsuit) \quad \Phi(M_1 v) = M_2 \Phi(v) \quad \forall v \in F^3$$.

   (Indeed, $(\clubsuit)$ amounts to "checking on a generator". If $(\clubsuit)$ holds then for every natural number $i$ a straightforward induction argument shows for every $v \in F^3$ that $$\begin{align*} \Phi(\rho_1(i + 3\mathbb{Z})v) &= \Phi(\rho_1(1+3\mathbb{Z})^i v) \\ &= \Phi( {M_1}^i v) \\ &= {M_2}^i \Phi(v) \\ &= \rho_2(1 + 3\mathbb{Z})^i\Phi(v) \\ &= \rho_2(i+3\mathbb{Z})\Phi(v) \end{align*}$$ )

   In turn, it suffices to verify the $(\clubsuit)$ holds for the basis vectors $e_1,e_2,e_3$ for $V = F^3$.

   Since $e_1$ and $v_1$ are $1$-eigenvectors for $M_1$ resp. $M_2$, we have $$\Phi(M_1 e_1)= \Phi(e_1) = v_1 = M_2 v_1.$$

   Since $e_2$ and $v_3$ are $\zeta$-eigenvectors for $M_1$ resp. $M_2$, we have $$\Phi(M_1 e_2)= \Phi(\zeta e_2) = \zeta\Phi(e_2) = \zeta v_3 = M_2 v_3.$$

   Since $e_3$ and $v_2$ are $\zeta^2$-eigenvectors for $M_1$ resp. $M_2$, we have $$\Phi(M_1e_3)= \Phi(\zeta^2 e_3) = \zeta^2\Phi(e_3) = \zeta^2 v_2 = M_2v_2.$$ Thus $(\clubsuit)$ holds and the proof is complete.

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   Alternatively, note that the matrix of $\Phi$ in the standard basis is given by $$[\Phi] = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \zeta^2 & \zeta \\ 1 & \zeta & \zeta^2 \end{bmatrix}$$

   Now, to prove that $\Phi \circ \rho_1(g) = \rho_2(g) \circ \Phi$, it suffices to check that $M_2[\Phi] = [\Phi] M_1$ i.e. that $$$$\begin{bmatrix} 0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$ \cdot $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & \zeta^2 & \zeta \\ 1 & \zeta & \zeta^2 \end{bmatrix}$$

   $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & \zeta^2 & \zeta \\ 1 & \zeta & \zeta^2 \end{bmatrix}$$

   \cdot $$\begin{bmatrix} 1 & 0 & 0\\ 0 & \zeta & 0 \\ 0 & 0 & \zeta^2 \end{bmatrix}$$ $$

   IN fact, both products yield the matrix $$\begin{bmatrix} 1 & \zeta & \zeta^2 \\ 1 & 1 & 1 \\ 1 & \zeta^2 & \zeta \end{bmatrix}$$ :::

4. Let $V$ be a $n$ dimensional $F$-vector space for $n \in \mathbb{N}$.

   Let $\operatorname{GL}(V)$ denote the group $$\operatorname{GL}(V) = \{ \text{all invertible $F$-linear transformations$\phi:V \to VParseError: KaTeX parse error: Expected 'EOF', got '}' at position 1: }̲\}$ where the group operation is composition of linear transformations.

   Recall that $\operatorname{GL}_n(F)$ denotes the group of all invertible $n \times n$ matrices.

   If $\mathcal{B} = \{b_1,b_2,\cdots,b_n\}$ is a choice of basis, show that the assignment $\phi \mapsto [\phi]_{\mathcal{B}}$ determines an isomorphism $$\operatorname{GL}(V) \xrightarrow{\sim} \operatorname{GL}_n(F).$$

   Here $[\phi]_{\mathcal{B}} = [M_{ij}]$ denotes the matrix of $\phi$ in the basis $\mathcal{B}$ defined by equations

   $$\phi(b_i) = \sum_{k=1}^n M_{ki} b_k.$$

   ::: {.solution} Lets write $\Phi$ for the mapping $$\Phi:\operatorname{GL}(V) \to \operatorname{GL}_n(F)$$ defined above.

   An important property -- proved in Linear Algebra -- is that for $\phi,\psi:V \to V$ we have $$(\heartsuit) \quad [\phi \circ \psi]_{\mathcal{B}} = [\phi]_{\mathcal{B}} \cdot [\psi]_{\mathcal{B}}.$$ In words: "once you choose a basis, composition of linear transformations corresponds to multiplication of the corresponding matrices".

   Now, since the matrix of the endomorphism $\phi:V \to V$ is equal to the identity matrix $\mathbf{I}_n$ if and only if $\phi =\operatorname{id}_V$, $(\heartsuit)$ shows at once that a linear transformation $\phi:V \to V$ is invertible if and only if $[\phi]_{\mathcal{B}}$ is an invertible matrix.

   This confirms that $\Phi$ is indeed a group homomorphism.

   To show that $\Phi$ is an isomorphism, we exhibit its inverse. Namely, we defined a group homomorphism $$\Psi:\operatorname{GL}_n(F) \to \operatorname{GL}(V)$$ and check that $\Psi$ is the inverse to $\Phi$.

   TO define $\Psi$, we introduce the linear isomorphism $\beta:F^n \to V$ defined by the rule $$\beta \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} =\sum_{i=1}^n a_i b_i.$$

   For an invertible matrix $M$, we define $$\Psi(M):V \to V$$ by the rule $$\Psi(M)(v) = \beta M \cdot \beta^{-1} v$$

   If $M_1,M_2 \in\operatorname{GL}_n(F)$ then for every $v \in V$ we have $$\begin{align*} \Psi(M_1M_2)v &= \beta M_1 M_2 \cdot \beta^{-1} v \\ &= \beta M_1 \beta^{-1} \beta M_2 \cdot \beta^{-1} v \\ &= \Psi(M_1)\Psi(M_2)v \end{align*}$$ This confirms that $\Psi$ is a group homomorphism.

   It remains to observe that for $M \in \operatorname{GL}_n(F)$ we have $$\Phi \circ \Psi (M) = M,$$ which amounts to the fact that $M$ is the matrix of $\Psi(M)$, and we must observe for $g \in \operatorname{GL}(V)$ hat $$\Psi \circ \Phi (g) = g$$ which amounts to the observation that the transformation $g:V \to V$ is determined by its effect on the basis vectors $b_i$ and hence by the matrix $\Phi(g)$.

   :::