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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.

ຕົວຢ່າງ​ທີ່​ໄດ້​ເຮັດ · 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]

ຂັ້ນ​ຕອນ​ຕໍ່​ຂັ້ນ​ຕອນ

  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.

ເປີດ​ເຜີຍ​ຄຳ​ຕອບ
\lambda = 5,\; \lambda = 2

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ວິທີການ: 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.

ຄໍາຖາມທີ່ຄົນຖາມ

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.

ເພີ່ມເຕີມໃນ Linear Algebra

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