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     11 <div id="Eigenvectors &amp; eigenvalues"><h2 id="Eigenvectors &amp; eigenvalues">Eigenvectors &amp; eigenvalues</h2></div>
     12 <p>
     13 let A be n × n, \(x \in \Re^n\) is an eigenvector of A if x ≠ 0 and \(\exists \lambda \in \Re\) such that \(Ax = \lambda x\) 
     14 </p>
     15 
     16 <p>
     17 x is eigenvector with corresponding eigenvalue λ.
     18 </p>
     19 
     20 <p>
     21 Is a given vector \(u \in \Re^n\) an eigenvector of a given A (n × n)?
     22 </p>
     23 <ul>
     24 <li>
     25 Do \(Au\), check if result is a multiple of u.
     26 
     27 </ul>
     28 
     29 <p>
     30 Is a given λ an eigenvalue of A?
     31 </p>
     32 <ul>
     33 <li>
     34 \(\exists x \ne 0\) such that \(Ax - \lambda x = 0 \leftrightarrow (A-\lambda I_n)x = 0\) with nontrivial solutions.
     35 
     36 </ul>
     37 
     38 <p>
     39 The solution set of \((A-\lambda I_n)x = 0\) is the eigenspace corresponding to λ.
     40 </p>
     41 
     42 <p>
     43 How to find a basis for the eigenspace of a given λ?
     44 </p>
     45 <ol>
     46 <li>
     47 calculate matrix for \(A-\lambda I_n\) where n is the number of rows or columns of A
     48 
     49 <li>
     50 reduce matrix to reduced echelon form
     51 
     52 <li>
     53 express solutions in parametric form (basic variables in terms of free variables)
     54 
     55 <li>
     56 basis for eigenspace is the set of the coefficients
     57 
     58 </ol>
     59 
     60 <p>
     61 If λ = 0, then Ax = 0 has a nontrivial solution (and A is <em>not</em> invertible).
     62 </p>
     63 
     64 <p>
     65 Eigenvectors corresponding to distinct eigenvalues are linearly independent.
     66 </p>
     67 
     68 <div id="Eigenvectors &amp; eigenvalues-Determinant"><h3 id="Determinant">Determinant</h3></div>
     69 <p>
     70 Geometric interpretation: let \(A = [a_1 \; a_2]\). then the determinant (absolute value) is the surface area (or volume in 3D):
     71 </p>
     72 
     73 <p>
     74 <img src="img/determinant-geometric-diagram.png" alt="Determinant geometric diagram" />
     75 </p>
     76 
     77 <p>
     78 Let A (n × n). A ~ U without scaling and using <em>r</em> row interchanges. then \(\det A = (-1)^r u_{11} \times \dots \times u_{nn}\)
     79 </p>
     80 
     81 <p>
     82 A is invertible iff \(\det A \ne 0\)
     83 </p>
     84 
     85 <p>
     86 \(\det AB = (\det A)(\det B)\)
     87 </p>
     88 
     89 <p>
     90 λ is an eigenvalue of A iff \(\det (A-\lambda I) = 0\) (the characteristic equation of A)
     91 </p>
     92 
     93 <p>
     94 The eigenvalues of A (n × n) are the solutions for λ. Multiplicity is the number of solutions for λ.
     95 </p>
     96 
     97 <div id="Eigenvectors &amp; eigenvalues-Similarity"><h3 id="Similarity">Similarity</h3></div>
     98 <p>
     99 given A and B (n × n), A is similar to B if ∃p s.t. \(A = PBP^{-1}\)
    100 </p>
    101 
    102 <p>
    103 If A and B are similar, then they have the same characteristic polynomials (and the same eigenvalues with the same multiplicities)
    104 </p>
    105 
    106 <div id="Eigenvectors &amp; eigenvalues-Diagonalization"><h3 id="Diagonalization">Diagonalization</h3></div>
    107 <p>
    108 A is diagonalizable if A is similar to a diagonal matrix.
    109 </p>
    110 
    111 <p>
    112 Diagonalization Theorem: A (n × n) is diagonalizable iff A has n linearly independent eigenvectors (the eigenbasis for \(\Re^n\))
    113 </p>
    114 
    115 <p>
    116 \(A = P D P^{-1} \leftrightarrow\) columns of P are linearly independent eigenvectors, and the diagonal values of D are the eigenvalues corresponding to the eigenvectors in P.
    117 </p>
    118 
    119 <p>
    120 How to diagonalize a matrix:
    121 </p>
    122 <ol>
    123 <li>
    124 Find eigenvalues of A
    125 
    126 <li>
    127 Find n = λ linearly independent eigenvectors
    128 
    129 <li>
    130 Construct \(P = \begin{bmatrix} p_1 &amp; p_2 &amp; \ldots &amp; p_n \end{bmatrix}\)
    131 
    132 <li>
    133 Construct D from the corresponding eigenvalues on the diagonal. Order of eigenvalues must match the order for columns of P.
    134 
    135 <li>
    136 Check \(A = p D p^{-1} \leftrightarrow Ap = pD\) (if p is invertible)
    137 
    138 </ol>
    139 
    140 <p>
    141 If A (n × n) has n distinct eigenvalues, it is diagonalizable.
    142 </p>
    143 
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