Free Online Flashcard Deck

7 Eigenvalues and Eigenvectors Free Online FlashCards

Study 7 Eigenvalues and Eigenvectors with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

What is an eigenvector?

Back

A nonzero vector satisfying Av=λvAv=\lambda v for a scalar λ\lambda. The scalar λ\lambda is the corresponding eigenvalue.

02
Front

Why is the zero vector not an eigenvector?

Back

The zero vector is excluded because A0=λ0A0=\lambda 0 holds for every scalar λ\lambda, so it cannot identify a specific eigenvalue.

03
Front

What does a negative eigenvalue indicate?

Back

If λ<0\lambda<0, the transformation reverses the eigenvector’s direction as well as scaling it.

04
Front

When is λ\lambda an eigenvalue of AA?

Back

λ\lambda is an eigenvalue exactly when A−λIA-\lambda I has a nontrivial nullspace, equivalently when det⁡(A−λI)=0\det(A-\lambda I)=0.

05
Front

How does the characteristic polynomial find eigenvalues?

Back

The characteristic polynomial is commonly pA(λ)=det⁡(λI−A)p_A(\lambda)=\det(\lambda I-A). Its roots are the eigenvalues of AA.

06
Front

How are eigenvectors found after eigenvalues?

Back

For each eigenvalue λ\lambda, solve (A−λI)v=0(A-\lambda I)v=0. All nonzero vectors in that nullspace are eigenvectors for λ\lambda.

07
Front

What is the eigenspace EλE_\lambda?

Back

The eigenspace is Eλ=ker⁡(A−λI)={v:Av=λv}E_\lambda=\ker(A-\lambda I)=\{v:Av=\lambda v\}. It contains the zero vector and all eigenvectors for λ\lambda.

08
Front

What is the difference between algebraic and geometric multiplicity?

Back

Algebraic multiplicity counts how often λ\lambda is a root of the characteristic polynomial; geometric multiplicity is dim⁡Eλ\dim E_\lambda.

09
Front

What is true of eigenvectors for distinct eigenvalues?

Back

Eigenvectors corresponding to distinct eigenvalues are linearly independent.

10
Front

What does it mean for a matrix to be diagonalizable?

Back

A matrix is diagonalizable if there are an invertible PP and a diagonal DD such that A=PDP−1A=PDP^{-1}.

11
Front

What do PP and DD contain in A=PDP−1A=PDP^{-1}?

Back

The columns of PP are eigenvectors of AA, and the corresponding eigenvalues appear in the same order on the diagonal of DD.

12
Front

How does diagonalization simplify matrix powers?

Back

If A=PDP−1A=PDP^{-1}, then Ak=PDkP−1A^k=PD^kP^{-1}, where Dk=diag⁡(λ1k,…,λnk)D^k=\operatorname{diag}(\lambda_1^k,\ldots,\lambda_n^k).