1.10

1. Consider the complex vector space \(M_2(\mathbb{C})\) and \(B = \begin{pmatrix} 1 & -1 \\ -4 & 4 \end{pmatrix} \in M_2(\mathbb{R})\)​

   Let \(T: M_2(\mathbb{C}) \to M_2(\mathbb{C})\), \(T(A) = BA\).

   a. Find \(\text{Null}(T)\) and \(\text{Range}(T)\). Verify the dimension theorem.

   Let’s find \([T]_C\) for \(C\), the canonical basis of \(M_2(\mathbb{C})\). Say \(e=\{E_{1},E_{2},E_{3},E_{4}\}\).

   \(T(E_1) = B E_1 = \begin{pmatrix} 1 & -1 \\ -4 & 4 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -4 & 0 \end{pmatrix}.\)

   \(T(E_2) = \begin{pmatrix} 0 & 1 \\ 0 & -4 \end{pmatrix}, \quad T(E_3) = \begin{pmatrix} -1 & 0 \\ 4 & 0 \end{pmatrix}, \quad T(E_4) = \begin{pmatrix} 0 & -1 \\ 0 & 4 \end{pmatrix}.\)​

   Therefore, \([T]_C = \begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ -4 & 0 & 4 & 0 \\ 0 & -4 & 0 & 4 \end{pmatrix}.\)​

   \(\text{Null}(T)\): We must find the solutions to the homogeneous system associated to \([T]_C\). So, we must row-reduce \([T]_C\).

   You can check we will get \(\begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \implies \begin{cases} x = z \\ y = w \end{cases}.\)

   Since these represent coordinates in \(E\), we get \(\text{Null}(T) = \langle \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \rangle.\)​

   \(\text{Range}(T)\): The \(\text{Range}(T)\) is everything spanned by the columns of \([T]_C\).

   So to find \(\text{Range}(T)\), it is enough to find the row-span of \([T]_C^T\).

   \([T]_C^T = \begin{pmatrix} 1 & 0 & -4 & 0 \\ 0 & 1 & 0 & -4 \\ -1 & 0 & 4 & 0 \\ 0 & -1 & 0 & 4 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -4 & 0 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.\)​

   (Row reductions: \(R_3 + R_1\), \(R_4 + R_2\)).

   The non-zero rows correspond to the basis of \(\text{Range}(T)\).

   \(\implies (1, 0, -4, 0) \text{ and } (0, 1, 0, -4) \text{ are the coordinates in } E \text{ of the matrices which span } \text{Range}(T).\)

   \(\implies \text{Range}(T) = \left\langle \begin{pmatrix} 1 & 0 \\ -4 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & -4 \end{pmatrix} \right\rangle\quad\quad \dim(\text{Range}(T)) = 2.\)​

   b. Describe \(T^2\).

   We have that \(T(A) = BA\)​\(\implies T^2(A) = T(T(A)) = T(BA) = B(BA) = B^2A.\)

2. Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) such that \(\det(A) = -3\).

   Calculate \(\det \begin{pmatrix} 2 & -2 & 0 \\ c+1 & -1 & 2a \\ d-2 & 2 & 2b \end{pmatrix}.\)​

   Way 1: Direct calculation.

   \(\det=-4\det(A)=12\)​

   Way 2: By applying row operations.

   \(\det \begin{pmatrix} 2 & -2 & 0 \\ c+1 & -1 & 2a \\ d-2 & 2 & 2b \end{pmatrix} \overset{\det(A^T) = \det(A)}{\longrightarrow} \det \begin{pmatrix} 2 & c+1 & d-2 \\ -2 & -1 & 2 \\ 0 & 2a & 2b \end{pmatrix}\\\overset{R_2+R_1}{\longrightarrow} \det \begin{pmatrix} 2 & c+1 & d-2 \\ 0 & c & d \\ 0 & 2a & 2b \end{pmatrix}\overset{R_3/2}{\longrightarrow} \det \begin{pmatrix} 2 & c+1 & d-2 \\ 0 & c & d \\ 0 & a & b \end{pmatrix}.\)

   Cofactor first column \(2 \cdot 2 \cdot \det \begin{pmatrix} c & d \\ a & b \end{pmatrix} = 4(-1) \cdot \det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = -4 \cdot \det(A) = 12.\)​

3. Prove that there exists a unique linear map \(T: \mathbb{R}^3 \to \mathbb{R}^3\) such that \((1, 1, 1)\) and \((-1, 1, 0)\) are eigenvectors with eigenvalue \(2\), and \((0, -2, 1)\) is an eigenvector with eigenvalue \(1\).

   For this \(T\), find \(\det(T)\) and \([T]_C\), where \(C\) is the canonical basis.

   Note that \(B = \{(1, 1, 1), (-1, 1, 0), (0, -2, 1)\}\), where \(\mathbf{v}_1 = (1, 1, 1)\), \(\mathbf{v}_2 = (-1, 1, 0)\), and \(\mathbf{v}_3 = (0, -2, 1)\)​ is a basis of \(\mathbb{R}^3\) since we have 3 linearly independent vectors in \(\mathbb{R}^3\).

   (You can verify this by checking that \(\begin{pmatrix} \mathbf{v}_{1} & \mathbf{v}_{2} & \mathbf{v}_{3} \end{pmatrix}\) is invertible.

   So we can define \(T: \mathbb{R}^3 \to \mathbb{R}^3\) by \(T(\mathbf{v}_1) = 2\mathbf{v}_1, \quad T(\mathbf{v}_2) = 2\mathbf{v}_2, \quad T(\mathbf{v}_3) = \mathbf{v}_3.\)​

   Since we have defined \(T\) on a basis, this is the only possible linear map satisfying the conditions, since if \(S: \mathbb{R}^3 \to \mathbb{R}^3\) has the same conditions...

   Then, since \(\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}\) is a basis, for any \(\mathbf{v} \in \mathbb{R}^3\), I can write \(\mathbf{v} = a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + a_3 \mathbf{v}_3\).

   \[ \implies S(\mathbf{v}) = a_1 S(\mathbf{v}_1) + a_2 S(\mathbf{v}_2) + a_3 S(\mathbf{v}_3) = a_1 T(\mathbf{v}_1) + a_2 T(\mathbf{v}_2) + a_3 T(\mathbf{v}_3) = T(\mathbf{v}). \]
   \[ \implies S(\mathbf{v}) = T(\mathbf{v}) \quad \forall \mathbf{v} \in V \implies S = T. \]

   So \(T\) is unique.

   \(\det(T)\): Since \(B = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}\) is a basis of eigenvectors of \(T\),

   \[ \implies T \text{ is diagonalizable with } [T]_B = \begin{pmatrix} [T(\mathbf{v}_1)]_B & [T(\mathbf{v}_2)]_B & [T(\mathbf{v}_3)]_B \end{pmatrix}. \]
   \[ \implies[T]_{B}= \begin{pmatrix} 2\mathbf{v}_{1} & 2\mathbf{v}_{2} & \mathbf{v}_{3} \end{pmatrix}_{B}\implies[T]_{B}= \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}\implies \det(T) = 2 \cdot 2 \cdot 1 = 4. \]
   \[ [T]_{C}:\text{ We need the change of basis matrix}\quad P_C^B: \quad [T]_C = (P_C^B)^{-1} [T]_B P_C^B= P_B^C [T]_B P_C^B. \]

   We have that \(P_B^C = \begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3\end{pmatrix} = \begin{pmatrix} 1 & -1 & 0 \\ 1 & 1 & -2 \\ 1 & 0 & 1 \end{pmatrix}.\)​

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