\(F\) denotes an algebraically closed field of characteristic 0. If you like, you can suppose that \(F = \mathbf{C}\) is the field of complex numbers.
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:
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.
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 \[Te_i = \left \{ \begin{matrix} e_{i+1} & i < n \\ e_1 & i =n \end{matrix} \right ..\]
Show that \(T\) is invertible and that \(T^n = \operatorname{id}_V\).
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 \(T^{n-1}\).
Consider the vector \(v_0 = \displaystyle \sum_{i=1}^n e_i\). Show that \(v_0\) is a \(1\)-eigenvector for \(T\).
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 \(\zeta = e^{2\pi i/n}\)).
Let \(v_1 = \displaystyle \sum_{i=1}^n \zeta^i e_i\). Show that \(v_1\) is a \(\zeta^{-1}\)-eigenvector for \(T\).
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.
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 \(T\).
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.
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?
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 \(M\)).
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 \(M_2\).
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.
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 \[\begin{bmatrix} 1 & \zeta & \zeta^2 \\ 1 & 1 & 1 \\ 1 & \zeta^2 & \zeta \end{bmatrix} \]
Let \(V\) be a \(n\) dimensional \(F\)-vector space for \(n \in \mathbb{N}\).
Let \(\operatorname{GL}(V)\) denote the group \[\operatorname{GL}(V) = \{ \text{all invertible $F$-linear transformations $\phi:V \to V$}\}\] 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
\[\phi(b_i) = \sum_{k=1}^n M_{ki} b_k.\]
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 \(\Phi(g)\).