added eigenvalues
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Robert
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@ -13,6 +13,305 @@
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\[
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\set[Av = \lambda v]{v \in \field^n} =: E_{\lambda}
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\]
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eigenspace belonging to $\lambda$
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eigenspace belonging to $\lambda$.
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\end{defi}
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\begin{eg}
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Let
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\[
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A = \begin{pmatrix}
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2 & 1 & -1 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{pmatrix}
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\]
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Then
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\begin{align*}
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A \cdot \begin{pmatrix}
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1 \\ 0 \\ 0
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\end{pmatrix}
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&= \begin{pmatrix}
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2 \\ 0 \\ 0
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\end{pmatrix}
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= 2 \cdot \begin{pmatrix}
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1 \\ 0 \\ 0
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\end{pmatrix} \\
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A \cdot \begin{pmatrix}
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1 \\ -1 \\ 0
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\end{pmatrix}
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&= \begin{pmatrix}
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1 \\ -1 \\ 0
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\end{pmatrix}
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= 1 \cdot \begin{pmatrix}
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1 \\ -1 \\ 0
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\end{pmatrix} \\
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A \cdot \begin{pmatrix}
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1 \\ 0 \\ 1
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\end{pmatrix}
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&= \begin{pmatrix}
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1 \\ 0 \\ 1
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\end{pmatrix}
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= 1 \cdot \begin{pmatrix}
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1 \\ 0 \\ 1
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\end{pmatrix}
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\end{align*}
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The eigenspaces are
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\begin{align*}
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E_2 &= \set[\kappa \in \realn]{\kappa \cdot \begin{pmatrix}1\\0\\0\end{pmatrix}} \\
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E_1 &= \set[\kappa, \rho \in \realn]{\kappa \cdot \begin{pmatrix}1\\-1\\0\end{pmatrix} + \rho \cdot \begin{pmatrix}1\\0\\1\end{pmatrix}}
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= \spn\set{\begin{pmatrix}1\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}}
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\end{align*}
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\end{eg}
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\begin{rem}
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The eigenspace to an eigenvalue $\lambda$ is a linear subspace.
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\end{rem}
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\begin{rem}
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We want to find $\lambda \in \field$, $v \in \field^n$ such that
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\[
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Av = \lambda v \iff (\underbrace{A - \lambda I}_{\in \field^{n \times n}}) v = 0
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\]
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If $(A - \lambda I)$ is invertible, then $v = 0$. So the interesting case is when $(A - \lambda I)$ not invertible.
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\[
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(A - \lambda I) \text{ not invertible} \iff \det(A - \lambda I) = 0
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\]
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This determinant is called the characteristic polynomial. This polynomial has degree $n$, and the eigenvalues are the roots of that polynomial.
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So let $\lambda$ be an eigenvalue of $A$, then
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\[
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(A - \lambda I) v = 0
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\]
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is a linear equation system for the components of $v$.
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\end{rem}
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\begin{eg}
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Let
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\[
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A = \begin{pmatrix}
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0 & 1 \\ -1 & 0
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\end{pmatrix}
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\in \cmpln^{2 \times 2}
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\]
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The characteristic polynomial is
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\[
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\det(A - \lambda I) = \begin{vmatrix}
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-\lambda & 1 \\ -1 & -\lambda
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\end{vmatrix}
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= \lambda^2 + 1
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\]
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Its roots are
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\begin{multicols}{2}
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\noindent
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\[ \lambda_1 = i \]
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\[ \lambda_2 = -i \]
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\end{multicols}
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\noindent To find the eigenvector belonging to $\lambda_1$, we declare $v_1 = (x, y) \in \cmpln^2$ and solve the linear equation system
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\begin{multicols}{2}\noindent
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\[ (A - \lambda_1 I) v_1 = 0 \]
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\begin{align*}
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-ix + 1y &= 0 \\
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-1x - iy &= 0 \\
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\end{align*}
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\end{multicols}
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\noindent It has the solutions $x = -i$ and $y = 1$, so
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\[
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v_1 = \begin{pmatrix}
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-i \\ 1
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\end{pmatrix}
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\]
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Doing the same for $v_2$ yields
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\[
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v_2 = \begin{pmatrix}
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i \\ 1
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\end{pmatrix}
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\]
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It is to be noted that the eigenvectors aren't unique (multiples of eigenvectors are also eigenvectors).
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\end{eg}
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\begin{eg}
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Let $D$ be a diagonal matrix, with the diagonal entries $\lambda_j$. Then
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\[
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\det(D - \lambda I) = \begin{vmatrix}
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\lambda_1 - \lambda & & & \\
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& \lambda_2 - \lambda & & \\
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& & \ddots & \\
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& & & \lambda_n - \lambda \\
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\end{vmatrix}
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\]
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The roots (eigenvalues) are $\lambda_1, \lambda_2, \cdots, \lambda_n$, and the eigenvectors are $De_i = \lambda_i e_i$.
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\end{eg}
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\begin{defi}
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$A \in \field^{n \times n}$ is called diagonalizable if there exists a basis of $\field^n$ that consists of eigenvectors.
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\end{defi}
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\begin{thm}
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A matrix $A \in \field^{n \times n}$ is diagonalizable, if and only if there exists a diagonal matrix $D$ and a invertible matrix $T$ such that
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\[
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D = \inv{T}AT
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\]
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\end{thm}
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\begin{proof}
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Let $e_1, e_2, \cdots, e_n$ be the canonical basis of $\field^n$. Define $TD\inv{T} = A$, and let $\lambda_1, \cdots, \lambda_n$ be the diagonal entries of $D$.
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Then we know that
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\begin{equation}
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De_i = \lambda_ie_i, ~~\forall i \in \set{1, \cdots n}
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\end{equation}
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Since $T$ is invertible, the $Te_1, \cdots Te_n$ form a basis.
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\begin{equation}
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A(Te_i) = T(\inv{T}AT)e_i = TDe_i = T\lambda_i e_i = \lambda_i (Te_i)
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\end{equation}
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Therefore $Te_i$ is an eigenvector of $A$ to the eigenvalue $\lambda_i$. Now let $v_1, \cdots, v_n$ be a basis of $\field^n$ and
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\begin{equation}
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Av_i = \lambda_iv_i, ~~\lambda_1, \cdots, \lambda_n \in \field^n
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\end{equation}
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Write write $v_1, \cdots, v_n$ as the columns of a matrix, therefore
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\begin{subequations}
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\begin{equation}
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T = (v_1, v_2, \cdots, v_n)
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\end{equation}
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\begin{equation}
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D = \begin{pmatrix}
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\lambda_1 & & \\
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& \vdots & \\
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& & \lambda_n
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\end{pmatrix}
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\end{equation}
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\end{subequations}
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So $Te_i = v_i$, and thus
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\begin{equation}
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A(Te_i) = Av_i = \lambda_iv_i = \lambda_i(Te_i) = T\lambda_ie_i = TDe_i
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\end{equation}
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This means that $(AT - TD)e_i = 0$, $\forall i \in \set{1, \cdots, n}$.
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\begin{equation}
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\implies AT = TD
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\end{equation}
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$T$ is invertible (left as an exercise for the reader), and thus
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\begin{equation}
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\implies \inv{T}AT = D
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\end{equation}
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\end{proof}
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\begin{eg}
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\begin{enumerate}[(i)]
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\item Let
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\[
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A = \begin{pmatrix}
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0 & 1 \\ -1 & 0
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\end{pmatrix}
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\]
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The eigenvalues and eigenvectors are
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\begin{multicols}{2}\noindent
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\[
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A \cdot \begin{pmatrix}
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-i \\ 1
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\end{pmatrix}
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=
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i \begin{pmatrix}
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-i \\ 1
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\end{pmatrix}
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\]
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\[
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A \cdot \begin{pmatrix}
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i \\ 1
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\end{pmatrix}
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=
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-i \begin{pmatrix}
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i \\ 1
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\end{pmatrix}
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\]
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\end{multicols}
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\noindent Therefore
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\[
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T = \begin{pmatrix}
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-i & i \\ 1 & 1
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\end{pmatrix}
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\]
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which has the inverse
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\[
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\inv{T} = \frac{1}{2} \begin{pmatrix}
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i & 1 \\ -i & 1
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\end{pmatrix}
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\]
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Finally,
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\[
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\inv{T}AT = \frac{1}{2}
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\begin{pmatrix}
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i & 1 \\ -i & 1
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\end{pmatrix}
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\begin{pmatrix}
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1 & 1 \\ i & -i
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\end{pmatrix}
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= \frac{1}{2}
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\begin{pmatrix}
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2i & 0 \\ 0 & -2i
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\end{pmatrix}
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=
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\begin{pmatrix}
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i & 0 \\ 0 & -i
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\end{pmatrix}
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\]
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This is a diagonal matrix, therefore $A$ is diagonalizable.
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\item The matrix
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\[
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\begin{pmatrix}
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0 & 1 \\ 0 & 0
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\end{pmatrix}
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\]
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is not diagonalizable since its only eigenvector is $(1, 0)^T$.
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\end{enumerate}
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\end{eg}
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\begin{rem}
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For diagonal matrices the following is true
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\[
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\begin{pmatrix}
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\lambda_1 & & & \\
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& \lambda_2 & & \\
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& & \ddots & \\
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& & & \lambda_3 \\
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\end{pmatrix}^k
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=
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\begin{pmatrix}
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\lambda_1^k & & & \\
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& \lambda_2^k & & \\
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& & \ddots & \\
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& & & \lambda_3^k \\
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\end{pmatrix}
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\]
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If $\inv{T}AT = D$ (where $D$ is a diagonal matrix), then
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\[
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D^k = (\inv{T}AT)^k = \underbrace{\inv{T}AT \cdot \inv{T}AT \cdot \cdots}_{k \text{ times}} = \inv{T}A^kT
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\]
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\[
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\implies A^k = TD^k\inv{T}
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\]
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\end{rem}
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\begin{thm}
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Let $A \in \realn^{n \times n}$ be a symmetric matrix, i.e. $A = A^T$. (Or if $A \in \cmpln^{n \times n}$ a self-adjoint matrix $A = A^H$).
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Then $A$ has an orthonormal basis consisting of eigenvectors and is diagonalizable.
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\end{thm}
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\begin{proof}
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Let $\lambda \in \cmpln$ be an eigenvalue of $A \in \field^{n \times n}$ with eigenvector $v \in \field^n$ and $A = A^H$. Then
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\begin{equation}
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\lambda\innerproduct{v}{v} = \innerproduct{v}{\lambda v} = \innerproduct{v}{Av} = \innerproduct{A^Hv}{v} = \innerproduct{Av}{v} = \innerproduct{\lambda v}{v} = \conj{\lambda} \innerproduct{v}{v}
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\end{equation}
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Therefore
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\begin{equation}
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(\lambda - \conj{\lambda})\underbrace{\innerproduct{v}{v}}_0 = 0
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\end{equation}
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\begin{equation}
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\implies (\lambda - \conj{\lambda}) = 0 \implies \lambda = \conj{\lambda} \implies \lambda \in \realn
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\end{equation}
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Now let $\lambda, \rho \in \realn$ be eigenvalues to the eigenvectors $v, w$, and require $\lambda \ne \rho$. Then
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\begin{equation}
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\rho\innerproduct{v}{w} = \innerproduct{v}{Aw} = \innerproduct{Av}{w} = \conj{\lambda}\innerproduct{v}{w} = \lambda\innerproduct{v}{w}
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\end{equation}
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And thus
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\begin{equation}
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\underbrace{(\rho - \lambda)}_{\ne 0} \underbrace{\innerproduct{v}{w}}_{=0} = 0 \implies v \perp w
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\end{equation}
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\end{proof}
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\end{document}
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