Created by M. Oki Orlandofrom the Noun Project

Eigenvector

Definition (Eigenvector)

For TL(V)T\in\mathscr{L}(V), an eigenvector of TT is a non-zero vector vVv\in V such that T(v)=λv for some λFT(v)=\lambda v\quad\text{ for some }\lambda \in \mathbb{F}

Definition (Eigenvalue)

The value λ\lambda is called an eigenvalue of TT.

Definition (Eigenspace)

If λ\lambda is an eigenvalue of TT, then the set of eigenvectors Wλ={vVT(v)=λv}W_\lambda=\{v\in V|T(v)=\lambda v\}is a subspace of VV, and we call WλW_\lambda the eigenspace corresponding to the eigenvalue λ.\lambda.

Remark

We can interpret the subspace WλW_\lambda as the kernel of the map TλIVT-\lambda I_V. If (TλIV)(v)=0(T-\lambda I_V)(v)=0, then we can rearrange this equation to get T(v)=λIV(v)=λvT(v)=\lambda I_V(v)=\lambda v. Therefore \mboxKer(TλIV)=Wλ\mbox{Ker}(T-\lambda I_V)=W_\lambda.

Theorem (Trace and determinant in terms of eigenvalues)

Let A={aij}\mathbf{A}=\{ a_{ij} \} be a k×kk\times k matrix of real elements with eigenvalues λ1,,λk\lambda_{1},\dots,\lambda_{k} (counting multiplicities) then can compute the trace using the eigenvalues Tr(A)=i=1kaii=i=1kλi\mathrm{Tr}(\mathbf{A})=\sum_{i=1}^{k}a_{ii}=\sum_{i=1}^{k}\lambda_{i}and the determinant det(A)=i=1kλi\det(\mathbf{A})=\prod_{i=1}^{k}\lambda_{i}

Theorem (Distinct eigenvalues have linearly independent eigenvectors)

For TL(V)T\in\mathscr{L}(V), let λ1,,λkF\lambda_1,\cdots,\lambda_k\in\mathbb{F} be distinct eigenvalues of TT with corresponding eigenvectors v1,,vkv_1,\cdots,v_k. The vectors v1,,vkv_1,\cdots,v_k are linearly independent.

Theorem (Operators on C\mathbb{C}-vector spaces have an eigenvalue)

If VV is a Finite Dimensional C\mathbb{C}-Vector Space and TL(V)T\in\mathscr{L}(V), then T has an eigenvalue or λ is an eigenvalue of T    vV, v0,\mboxs.t.(TλIV)(v)=0\lambda\text{ is an eigenvalue of }T\iff\exists v\in V, \ v\not=0, \mbox{ s.t. }(T-\lambda I_V)(v)=0

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