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Diagonalisation and matrix powers

A = PDP⁻¹, computing Aⁿ, and the matrices that cannot be diagonalised.

With a full set of eigenvectors, A = PDP⁻¹ where D is diagonal, so Aⁿ = PDⁿP⁻¹: Fibonacci in closed form, Markov chains at equilibrium. Picture it: the transformation stretching along eigenvector directions and nothing else. Think it: symmetric matrices always diagonalise with orthogonal eigenvectors (the spectral theorem); defective ones need Jordan form.

ឧទាហរណ៍​ដែល​បាន​ធ្វើ · eigenvalues of [[4,1],[2,3]]

Eigenvalues of [[4,1],[2,3]]

\left[\begin{matrix}4 & 1\\2 & 3\end{matrix}\right]

ជំហាន​ដោយ​ជំហាន

  1. \det(A - \lambda I) = 0

    Eigenvalues are the roots of the characteristic polynomial.

  2. \det\left[\begin{matrix}4 - \lambda & 1\\2 & 3 - \lambda\end{matrix}\right] = 0

    Subtract λ from the diagonal.

  3. \lambda^{2} - 7 \lambda + 10 = 0

    Expand the determinant.

  4. \left(\lambda - 5\right) \left(\lambda - 2\right) = 0

    Factor.

  5. \lambda = 5, \lambda = 2

    Eigenvalues (with multiplicity).

  6. \lambda = 2:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}- \frac{1}{2}\\1\end{matrix}\right]

    Solve (A − 2I)v = 0 for a basis eigenvector.

  7. \lambda = 5:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}1\\1\end{matrix}\right]

    Solve (A − 5I)v = 0 for a basis eigenvector.

បង្ហាញ​ចម្លើយ
\lambda = 5,\; \lambda = 2

ឥឡូវ​អ្នក ជ្រើស​បញ្ហា ឬ​វាយ​បញ្ចូល ឬ​គូរ​របស់​អ្នក​ផ្ទាល់ ។ ជំហាន​នីមួយៗ រូបភាព ចម្លើយ​ត្រូវ​បាន​លាក់​រហូត​ដល់​អ្នក​សួរ ។

រក្សា​ការ​ធ្វើការ​របស់​អ្នក​ផ្ទាល់

គណនី​ឥត​គិត​ថ្លៃ​បន្ថែម​ចំណាំ​លើ​មេរៀន​នីមួយៗ កំណត់​ហេតុ​នៃ​អ្វី​ដែល​អ្នក​បាន​បញ្ចប់ បញ្ហា​ដែល​អ្នក​បាន​ដោះស្រាយ​នៅ​កន្លែង​មួយ និង​គ្រូ​ដែល​អ្នក​អាច​សួរ​អំពី​ទំព័រ​នេះ ។ គណិតវិទ្យា​ដោយ​ខ្លួន​វា​គឺ​បើក​ចំហ​សម្រាប់​អ្នក​ទាំងអស់​គ្នា ចុះឈ្មោះ​ចូល ឬ​មិន​ចុះឈ្មោះ​ចូល ។

ចុះឈ្មោះ ចូល

និមិត្ត​សញ្ញា​ដែល​បាន​ប្រើ​នៅ​ទីនេះ

ចុច​និមិត្ត​សញ្ញា​ណាមួយ​សម្រាប់​ការ​កំណត់​ពេញលេញ រូបភាព និង​តួអក្សរ​នីមួយៗ​មាន​ន័យ​អ្វី ។

របៀប​: Diagonalisation and matrix powers

  1. Eigenvalues are the roots of the characteristic polynomial.
  2. Subtract λ from the diagonal.
  3. Expand the determinant.
  4. Factor.
  5. Eigenvalues (with multiplicity).
  6. Solve (A − 2I)v = 0 for a basis eigenvector.
  7. Solve (A − 5I)v = 0 for a basis eigenvector.

សំណួរ​ដែល​មនុស្ស​សួរ

What does a determinant mean geometrically?

It is the factor by which the matrix scales area (2×2) or volume (3×3), with a negative sign if orientation flips. Zero means the matrix flattens space and cannot be undone.

What is an eigenvector?

A direction the matrix does not turn: it only stretches it by the eigenvalue. Along eigenvectors a complicated matrix acts like multiplication by a number.

Why is matrix multiplication not commutative?

Because a matrix is a transformation and AB means "do B, then A". Rotating then reflecting is not the same as reflecting then rotating.

បន្ថែម​ទៀត​ក្នុង Linear Algebra