1-d invariant subspace — Jemoka Knowledge Base

eigenvalue is the scalar needed to scale the basis element of a one dimensional invariant subspace of a Linear Map to represent the behavior of the map:

\begin{equation} Tv = \lambda v \end{equation}

Note we require v \neq 0 because otherwise all scalars count. eigenvector is a vector that forms the basis list of length 1 of that 1-D invariant subspace under T. “operators own eigenvalues, eigenvalues own eigenvectors” Why is eigenvalue consistent per eigenvector? Because a linear map has to act on the same way to something’s basis as it does to the whole space. Motivation Take some subspace U \subset V:

\begin{equation} U = \{\lambda v\ |\ \lambda \in \mathbb{F}, v \in V\} = span(v) \end{equation}

Now, if T|_{U} is an operator on U, U would be an invariant subspace of T of dimension 1 (its basis being the list \{v\}). Therefore, for some vector v \in U (basically like various scalings of v), T will always send back to U so we can represent it yet again with another scalar on v, like \lambda v. In this case, then, we can write that:

\begin{equation} Tv = \lambda v \end{equation}

And then the usual definition of eigenvalues persist. constituents linear map T \in \mathcal{L}(V) vector v \in V, such that v \neq 0 scalar \lambda \in \mathbb{F} requirements If there exists v \in V such that v\neq 0 and:

\begin{equation} Tv = \lambda v \end{equation}

then, \lambda is called an eigenvalue, and v the eigenvector. additional information properties of eigenvalues Suppose V in finite-dimensional, T \in \mathcal{L}(V) and \lambda \in \mathbb{F}, then: \lambda is an eigenvalue of T T - \lambda I is not injective T - \lambda I is not surjective T - \lambda I is not invertable Showing one shows all. Proof: 1 \implies 2 Suppose \lambda is an eigenvalue of T. Then, we have some v \in V such that:

\begin{equation} Tv = \lambda v \end{equation}

Now:

\begin{align} &Tv = \lambda v \\ \Rightarrow\ & Tv - \lambda v = 0 \\ \Rightarrow\ & Tv - \lambda Iv = 0 \\ \Rightarrow\ & (T-\lambda I)v = 0 \end{align}

the last step by (T+S)v = Tv+Sv, the property of the vector space of \mathcal{L}(V) (or any \mathcal{L}). And therefore, v \in null\ (T-\lambda I), and v\neq 0. And so null\ (T-\lambda I) \neq \{0\} and so T-\lambda I is not injective, as desired. The reverse of this result shows the opposite direction that 1 \implies 2. The others I \in \mathcal{L}(V), T \in \mathcal{L}(V), \mathcal{L}(V) is closed, so (T - \lambda I) \in \mathcal{L}(V), and so it is an operator. Having 2) implies all other conditions of non-injectivity, non-surjectivity, non-invertiblility by injectivity is surjectivity in finite-dimensional operators list of eigenvectors are linearly independent Let T \in \mathcal{L}(V), suppose \lambda_{j} are distinct eigenvalues of T, and v_1, \ldots, v_{m} the corresponding eigenvectors, then v_1, \ldots, v_{m} is linearly independent. proof: We will show this by contradiction. Suppose v_1, \ldots, v_{m} are linearly dependent; then, by the Linear Dependence Lemma, \exists v_{j} such that:

\begin{equation} v_{j} \in span(v_1, \dots, v_{j-1}) \end{equation}

Meaning:

\begin{equation} v_{j} = a_1v_1 + \dots + a_{j-1}v_{j-1} \end{equation}

Given the list is a list of eigenvalues, we can apply T to both sides to get:

\begin{equation} \lambda_{j}v_{j} = a_1\lambda_{1}v_1 + \dots + a_{j-1}\lambda_{j-1}v_{j-1} \end{equation}

We can also get another definition for \lambda_{j} v_{j} by simply multiplying the definition for v_{j} above by \lambda_{j}:

\begin{align} &v_{j} = a_1v_1 + \dots + a_{j-1}v_{j-1}\ \text{from above} \\ \Rightarrow\ & \lambda_{j} v_{j} = a_1\lambda_{j}v_1 + \dots + a_{j-1}\lambda_{j}v_{j-1} \end{align}

Now, subtracting our two definitions of \lambda_{j} v_{j}, we get:

\begin{equation} 0 = a_1 (\lambda_{j} - \lambda_{1})v_{1} + \dots +a_{j-1} (\lambda_{j} - \lambda_{j-1})v_{j-1} \end{equation}

Recall now that the eigenvalue list \lambda_{j} are distinct. This means all \lambda_{j} - \lambda_{k \neq j} \neq 0. No v_{j} =0; so if we choose the smallest positive integer for j, the list before it v_1, \dots, v_{j-1} is linearly independent (as no value in that list would satisfy the Linear Dependence Lemma). This makes a_{j} =\dots =a_{j-1} = 0. And yet, substituting this back into the expression for v_{j}, we have v_{j} = 0, reaching contradiction. So therefore, the list of eigenvectors are linearly independent. \blacksquare operators on finite dimensional V has at most dim V eigenvalues As a corollary of the above result, suppose V is finite dimensional; then, each operator on V has at most dim\ V distinct eigenvalues because their eigenvectors form an linearly independent list and length of linearly-independent list \leq length of spanning list. eigenspaces are disjoint the eigenspaces of a Linear Map form a direct sum: proof: Corollary of result above. Because eigenvectors (i.e. bases) from distinct eigenspaces are linearly independent. So the only way to write 0 is by taking each to 0. So by taking the bases all to 0, you take the 0 vector from each space, which shows that the eigenspaces are a direct sum. \blacksquare finding eigenvalues with actual numbers \begin{equation} \lambda_{j} \in Spec(T) \Rightarrow det(\lambda_{j}I-T) = 0 \end{equation} The right polynomial det(\lambda_{j} I-T) = 0 is named the “characteristic polynomial.” natural choordinates of a map Given the eigenvectors (x+,y+), (x-,y-), we can change coordinates of your matrix into the natural choordinates.

\begin{equation} A = \begin{pmatrix} x+ & x- \\y+ & y- \end{pmatrix} \begin{pmatrix} \lambda+ & 0 \\ 0 & \lambda- \end{pmatrix} \begin{pmatrix} x+ & x- \\y+ & y- \end{pmatrix}^{-1} \end{equation}

This makes scaling matricides much much easier. If you think about multiplying the above matrix n times, the inverse and non-inverse cancells out. similar matrices Let A,B be defined:

\begin{equation} A = C B C^{-1} \end{equation}

and of course:

\begin{equation} B = C^{-1} B C \end{equation}

where, A,B,C \in \mathcal{L}(V) A, B has the same eigenvalues. invertable matricies Let T \in \mathcal{L}(V) be invertable. If \lambda is an eigenvalue of T, then \frac{1}{\lambda} is an eigenvalue of T. Furthermore, T and T^{-1} share eigenvectors with eigenvalues \lambda and \frac{1}{\lambda} symmetric matricies have a real basis of eigenvalues this falls out of the real spectral theorem.