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1.6

The characteristic polynomial

Definition

Given \(T\in \text{End}(V)\), the characteristic polynomial of \(T\) is \(P_T(x)=\det (T-\lambda I)\)​

Recall

  • \(P_{T}(x)=\det(T-\lambda I)=\det\left(A-\lambda I\right)\) where \(A=[T]_\mathcal{B}\)

  • The identity is not same, one is identity function(operators), another one is identity matrix

  • \(\lambda\) is a root of \(P_T(x)\) iff \(\lambda\) is an eigenvalue of \(T\)

  • \(P_T(x)\) leading coefficients is \(\pm 1\) of degree \(n=\dim V\)

Example

If \([T]_{\mathcal{B}}=\begin{pmatrix}\lambda_1I_{n_1} & & \\ & \ddots & \\ & & \lambda_{k}I_{nk}\end{pmatrix}\Rightarrow P_{T}\left(x\right)=\pm\left(x-\lambda_1\right)^{n_1}\cdots\left(x-\lambda_{k}\right)^{n_k}\) where each is block matrix

distinct

If \([T]_{\mathcal{B}}=\begin{pmatrix}\lambda_1 & & \star\\ & \ddots & \\ & & \lambda_{k}\end{pmatrix}\Rightarrow P_{T}\left(x\right)=\pm\left(x-\lambda_1\right)^{n_1}\cdots\left(x-\lambda_{k}\right)^{n_{k}}\)

Example

\(T: \mathbb{R}^2 \to \mathbb{R}^2\), \(T(e_1) = e_1\), \(T(e_2) = e_1 + e_2\)

\([T]=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\rightarrow P_{T}(x)=(x-1)^2\)​

Theorem

\(\mathbb{F}=\mathbb{C}\) \(T\) is diagonalizable iff \(P_{T}(x)=\pm\left(x-\lambda_1\right)^{n_1}\cdots\left(x-\lambda_{k}\right)^{n_{k}}\) and \(\dim V_{\lambda_{i}}=n_{i}\)

algebra multiplicity

Proof

\(\Rightarrow)\)

\(\Leftarrow)\) \(T\) is triangularizable if and only if \([T]=\begin{pmatrix}\lambda_1 & & & & \\ & \lambda_1 & & \star & \\ & & \ddots & & \\ & & & \lambda_{k} & \\ & & & & \lambda_{k}\end{pmatrix}\)

\(V_{\lambda_i} = \text{Null}(T - \lambda_i I)\), with \(\dim V_{\lambda_i} \leq n_i\). geometric multiplicity

If \(\dim V_{\lambda_i} = n_i\), then \(\dim V_{\lambda_1} + \cdots + \dim V_{\lambda_m} = n\) and \(T\) is diagonalizable.

An alternative proof

Every \(T\) is triangularizable in \(\mathbb{C}\)

Proof (Induction on \(n=\dim V\))

\(n=1 \checkmark\)

Given \(T\), let \(v_1\) be an eigenvector of \(T\) associated to the eigenvalue \(\lambda\))

Let \(U=\langle v_1\rangle\hookrightarrow V\), which is \(T\)-invariant

Now, consider \(W=V/U\), \(\dim W=\dim V-1=n-1\)​

Let \(\tilde{T}:W\to W\) defined by: \(\tilde{T}(v+U)=T(v)+U\)

Check: if \(v'+U=v+U\iff v'-v\in U\), then \(T(v'-v)\in U\iff T(v')+U=T(v)+U\)

By the inductive hypothesis, \(\tilde{T}\) is triangularziable, then \(\exists \tilde{\mathcal{B}}\) of \(W\), \(v_2+U,...,v_n+U\) (n-1)

Then \(\tilde{T}(v_{j}+U)\in\langle v_2+U,\ldots,v_{j}+U\rangle,\forall j=2,\ldots,n\)​

Now consider \(v_1,v_2,...,v_n\). Claim: \(v_1,v_2,...,v_n\) is a basis of \(V\) (\(\mathcal{B}\))

Finally, \([T]_\mathcal{B}\) is upper triangular. If fact: \(T(v_1)\in \lang v_1\rang\), since this, \(T(v_j)\in \lang v_2,...,v_j,v_1\rang\)​

Proof of claim

\(a_1 v_1 + a_2 v_2 + \cdots + a_n v_n = 0\)
\(\Rightarrow a_1 v_1 = -(a_2 v_2 + \cdots + a_n v_n)\)(in \(V\))
\(\Rightarrow 0 = -a_2 (v_2 + U) - \cdots - a_n (v_n + U)\) (in \(W\), after applying \(\pi\))
\(\Rightarrow a_2 = \cdots = a_n = 0, a_1 = 0\).

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