maths.free › Linear Algebra › Diagonalisation and matrix powers

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.

Apstrādāts piemērs · eigenvalues of [[4,1],[2,3]]

Linear-Algebra · Eigenvalues

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

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

Solis pēc kārtas

  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.

Atbildēt uz šo jautājumu
\lambda = 5,\; \lambda = 2

Tagad tu Izvēlieties problēmu, vai ierakstiet vai zīmējiet savu. Katrs solis, attēls, atbilde slēpta, līdz jūs lūgt.

Saglabāt savu darbu

Bezmaksas konts pievieno piezīmes par katru nodarbību, ierakstu par to, ko esat pabeidzis, jūsu atrisinātās problēmas vienā vietā, un pasniedzēju, kuru varat jautāt par šo lapu. Pati matemātika ir pieejama ikvienam, kas ir vai nav parakstījies.

Pierakstīties Pieteikšanās

Šeit izmantotie simboli

Piesitiet visiem simboliem, kas apzīmē pilnu definīciju, attēlu un ko katrs burts tajā nozīmē.

Kā lietot: 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.

Jautājumi, ko cilvēki vaicā

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.

Vairāk Linear Algebra

DeterminantsMatrix inverseEigenvalues and eigenvectorsMatrix multiplicationRow reductionVector spaces, span and linear independenceOrthogonality, projections and least squaresLinear transformations and change of basis