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1.3

Definition: Suppose \(T \in L(V, V)\) and denote \(A = [T]_B\) in any basis \(B\).

The characteristic polynomial of \(A\) is given by \(\chi_A(x) = \det(x \text{Id} - A)\).

Note: The roots of \(\chi_A(x)\) are the eigenvalues of \(T\).

  1. Find the eigenvalues and the associated eigenvectors of the matrix \(A = \begin{pmatrix} 0 & 3 \\ 3 & 0 \end{pmatrix}\)​

    Note that \(\chi_{A}(x)=\det(x\text{Id}-A)=\det\begin{pmatrix}x & -3\\ -3 & x\end{pmatrix}\).

    Then \(\chi_A(x) = x^2 - 9 = (x + 3)(x - 3)\). Thus the eigenvalues of \(A\) are \(3\) and \(-3\).

    • \(\lambda = 3\): \(\text{Null}(A - 3 \text{Id}) = \text{Null}\begin{pmatrix} -3 & 3 \\ 3 & -3 \end{pmatrix} = \text{Null}\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix}\) (row reducing) \(= \langle (1, 1) \rangle\).
    • \(\lambda = -3\): \(\text{Null}(A + 3 \text{Id}) = \text{Null}\begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix} = \text{Null}\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\) \(= \langle (-1, 1) \rangle\).

Definition: A linear map \(T \in L(V, V)\) is called triangulable if there exists some basis \(B\) of \(V\) such that \([T]_B\) is upper triangular.

Theorem: Let \(V\) be finite-dimensional and \(T \in L(V, V)\). \(T\) is triangulable \(\iff \chi_T(x) = (x - c_1)^{r_1} \cdots (x - c_k)^{r_k}\).

Note: By the Fundamental Theorem of Algebra, if \(V\) is a complex vector space (\(\mathbb{F} = \mathbb{C}\)), \(\chi_T(x) \in \mathbb{C}[x]\), and so we can always factor \(\chi_T\) as a product of linear terms.

\(\Rightarrow T\) is always triangulable.

Example: \(x^2 + 1 = (x - i)(x + i)\) in \(\mathbb{C}\) but not in \(\mathbb{R}\).

  1. Let \(T: \mathbb{R}^3 \to \mathbb{R}^3\), \(T(x, y, z) = [-3x + y, -x + z, 5x - 2y]\). Find a basis \(B\) of \(\mathbb{R}^3\) such that \([T]_B\) is upper triangular.

    Step 1: Find the eigenvalues of \(T\).

    So, consider \([T]_C = \begin{pmatrix} -3 & 1 & 0 \\ -1 & 0 & 1 \\ 5 & -2 & 0 \end{pmatrix}\), where \(C\) = canonical basis.

    \(\Rightarrow \chi_T(x) = \det(x \text{Id} - [T]_C) = \det\begin{pmatrix} x + 3 & -1 & 0 \\ -1 & x & -1 \\ 5 & -2 & x \end{pmatrix}\)

    \(= (x + 3)x^2 + 0 - 5 - 0 + 2(x + 3) + x = (x + 1)^3\)​

    \(\Rightarrow -1\) is the only eigenvalue of \(T\).

    Step 2: Choose an eigenvalue \(\lambda\) and find \(U = \text{Range}(T - \lambda \text{Id})\) (invariant subspace).

    We choose \(\lambda = -1\), and we have that \(T+\text{Id}=\begin{pmatrix}-2 & 1 & 0\\ -1 & 1 & 1\\ 5 & -2 & 1\end{pmatrix}\).

    To find the range, we transpose the matrix and row reduce:

    \((T+\text{Id})^{T}=\begin{pmatrix}-2 & -1 & 5\\ 1 & 1 & -2\\ 0 & 1 & 1\end{pmatrix}\to\begin{pmatrix}0 & 1 & 1\\ 1 & 0 & -3\\ 0 & 0 & 0\end{pmatrix}\)​

    \(\Rightarrow \text{Range}(T + \text{Id}) = \langle (0, 1, 1), (1, 0, -3) \rangle\).

    \(U = \langle (0, 1, 1), (1, 0, -3) \rangle\) with basis \(B_U = \{(0, 1, 1), (1, 0, -3)\} = \{u_1, u_2\}\).

    Step 3: Find \(T|_U : U \to U\) (the representation of \(T\) restricted to \(U\) in the basis \(B_U\)).

    Note that: \(T|_U(u_1) = (1, 1, -2) = u_1 + u_2\), \(T|_U(u_2) = (-3, -4, 5) = -4u_1 - 3u_2\).

    Thus \(\big[T|_U\big]_{B_U} = \begin{pmatrix} 1 & -4 \\ 1 & -3 \end{pmatrix}\).

    Step 4: Triangulate \(\big[T|_U\big]_{B_U}\) (so we need to find an eigenvalue of \(T|_U\)).

    \(\chi_{T|_U}(x) = \det(x \text{Id} - T|_U) = \det\begin{pmatrix} x - 1 & 4 \\ -1 & x + 3 \end{pmatrix}\) \(=(x+1)^2\)​

    \(\mu = -1\) is an eigenvalue of \(T|_U\), and the eigenspace is \(\text{Null}(T|_U + \text{Id}) = \text{Null}\begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix} = \text{Null}\begin{pmatrix} 1 & -2 \\ 0 & 0 \end{pmatrix} = \langle (2, 1) \rangle\). This is coordinate for my basis \(\mathcal{B}_U\)​

    An eigenvector is \(2u_1 + u_2 = (1, 2, -1)\)​

    So now I choose the following basis for \(U\): \(B_U' = \{2u_1 + u_2, u_1\}\) (eigenvector of \(T|_U\) associated to \(\mu = -1\)).

    \(\Rightarrow \big[T|_U\big]_{B_U'} = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}\) (upper triangular).

    Note: \(T|_U(2u_1 + u_2) = - (2u_1 + u_2)\) and \(T|_{U}(u_1)=u_1+u_2=1\cdot(2u_1+u_2)-u_1\)​

    Step 4: Extend \(B_U'\) to a basis of \(\mathbb{R}^3\): We take \(B = \{2u_1 + u_2, u_1, (1, 0, 0)\}\) where \(2u_1+u_2=(1,2,-1)\) and \(u_1=(0,1,1)\)​

    And now, \([T]_{B}=\begin{pmatrix}[T|_{U}]_{B_{U}^{\prime}} & [T(1,0,0)]_{B}\\0&0\end{pmatrix}=\begin{pmatrix}-1 & 1 & -2\\ 0 & -1 & 3\\ 0 & 0 & -1\end{pmatrix}\).

    Calculation for \(T(1, 0, 0)\)​ : \(T(1,0,0)=(-3,-1,5)=-2(1,2,-1)+3(0,1,1)-(1,0,0)\).

Definition: A linear map \(T \in L(V, V)\) is called diagonalizable if \(\big[T\big]_B\) is diagonal for some basis \(B\).

Theorem: \(T\) is diagonalizable \(\iff \chi_T(x) = (x - c_1)^{r_1} \cdots (x - c_k)^{r_k}\)​where \(r_{i}=\dim(\text{Null}T-c_{i}\text{Id}))\) (eigenspace).

Theorem: \(T \in L(V, V)\) is diagonalizable \(\iff\) there exists a basis \(B\) of \(V\) consisting of eigenvectors of \(T\).

  1. Show that \(T: \mathbb{R}^2 \to \mathbb{R}^2\), \(T(x, y) = (2y, 2x)\), is diagonalizable and find a matrix \(P\) such that \(P^{-1}[T]_C P\) is diagonal.

    We have that \(\big[T\big]_C = \begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix}\).

    \(\Rightarrow\chi_{T}(x)=\det\begin{pmatrix}x & -2\\ -2 & x\end{pmatrix}=x^2-4=(x-2)(x+2)\).

    \(\Rightarrow\) My eigenvalues are \(2\) and \(-2\).

    Let's see if we have a basis of \(\mathbb{R}^2\) of eigenvectors.

    For \(\lambda = 2\): \(\text{Null}(T - 2 \text{Id}) = \text{Null}\begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix} = \langle (1, 1) \rangle\).

    For \(\lambda = -2\): \(\text{Null}(T + 2 \text{Id}) = \text{Null}\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} = \langle (-1, 1) \rangle\).

    Now, \(\{(1, 1), (-1, 1)\}\) is my basis of eigenvectors of \(T\).

    \(\Rightarrow T\) is diagonalizable\(\big[T\big]_B = \begin{pmatrix} 2 & 0 \\ 0 & -2 \end{pmatrix}\).

    And \(P = P_B^C\) (change of basis matrix).

    \(\Rightarrow P^{-1} \big[T\big]_C P = \big[T\big]_B\) is a diagonal matrix.

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