Eigenspaces and Eigenbases

• (Friedberg 5.10) Let be a linear operator on and let be distinct eigenvalues of . If are the corresponding eigenvectors then are linearly independent.

• Let be an eigenvalue of , a linear operator over . The set

  is called the Eigenspace of corresponding to eigenvalue .

  For a matrix , the eigenspace of is the eigenspace of .

  The eigenspace is a subspace of .

• We define the geometric multiplicity as

  That is, it is the dimension of the eigenspace

• (Friedberg 5.12) Let be a linear operator on a finite dimensional vector space . If is an eigenvalue of , then .

  • Intuition: The geometric multiplicity is not necessarily the same as the algebraic multiplicity. By definition, an eigenvector must be nonzero. However, the zero vector may appear as a solution to the linear system corresponding to . Furthermore cannot exceed because that implies there are more than eigenvectors, that appears more than .

• (Friedberg Lem.5.13) Let be a linear operator on and be distinct eigenvalues of . For each let . If

  Then

  • Proof: The ’s form a linearly independent set. Therefore, the only non-trivial way to get is to set all .

• (Friedberg 5.13) Let be a linear operator on and be distinct eigenvalues of . Let be a finite linearly independent subset of . Then

  is a linearly independent subset of

• (Friedberg 5.14) Let be a linear operator on a finite vector space such that the characteristic polynomial of splits. Let be distinct eigenvalues of . Then

  • is diagonalizable if and only if the for all .
  • If is diagonalizable and is a basis for then
    forms a basis for called the eigenbasis. It contains eigenvectors of .

• (Friedberg 5.16) Let be a linear operator on a finite dimensional vector space . is diagonalizable if and only if is the direct sum of the eigenspaces of .

  • Suppose is diagonalizable. Each eigenspace is -invariant with characteristic polynomial . Which means by (Friedberg 5.29) The characteristic polynomial of is the product
    Where is the algebraic multiplicity of each eigenvalue, equal to the dimension of the corresponding eigenspace.

• If is a diagonalizable matrix, then it has an eigendecomposition which is obtained using the change of coordinate matrix

  Where is the matrix consisting of linearly independent eigenvalues and .

Spectral Theorem §

Links §

• Friedberg, Insel and Spence - Ch. 5