Vector subspace and eigenspace

A vector subspace is a subset of a vector space that is itself a vector space.

More formally, the vector subspace  of the vector space satisfies the following conditions:

1) Commutative and associative addition for all elements of the closed set.

\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}}=\boldsymbol{\mathit{v_{2}}}+\boldsymbol{\mathit{v_{1}}}

(\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}})+\boldsymbol{\mathit{v_{3}}}=\boldsymbol{\mathit{v_{1}}}+(\boldsymbol{\mathit{v_{2}}}+\boldsymbol{\mathit{v_{3}}})

2) Associativity and distributivity of scalar multiplication for all elements of the closed set

c_1(c_2\boldsymbol{\mathit{v_{1}}})=(c_1c_2)\boldsymbol{\mathit{v_{1}}}

c_1(\boldsymbol{\mathit{v_{1}}}+\boldsymbol{\mathit{v_{2}}})=c_1\boldsymbol{\mathit{v_{1}}}+c_1\boldsymbol{\mathit{v_{2}}}

(c_1+c_2)\boldsymbol{\mathit{v_{1}}}=c_1\boldsymbol{\mathit{v_{1}}}+c_2\boldsymbol{\mathit{v_{1}}}

where c_1 and c_2 are scalars.

3) Scalar multiplication identity.

\boldsymbol{\mathit{1}}\boldsymbol{\mathit{v_{1}}}=\boldsymbol{\mathit{v_{1}}}

4) Additive inverse.

\boldsymbol{\mathit{v_{1}}}+(-\boldsymbol{\mathit{v_{1}}})=0

5) Existence of null vector , such that

\boldsymbol{\mathit{0}}+\boldsymbol{\mathit{v_{1}}}=\boldsymbol{\mathit{v_{1}}}

6) Closed under addition: the sum of any two or more vectors in is another vector in .

7) Closed under scalar multiplication: the product of any vector in with a scalar is another vector in .

For example, in the  space, a plane through the origin is a subspace, as is a line through the origin. Hence, and are subspaces of . The entire space and the single point at the origin are also subspaces of the space. This implies that a subspace contains a set of orthonormal basis vectors.

An eigenspace is the set of all eigenvectors associated with a particular eigenvalue. In other words, it is a vector subspace formed by eigenvectors corresponding to the same eigenvalue. Consider the eigenvalue equation , where . If the eigenvalues of ,  and are , and respectively, then  and are the eigenspaces of the operator . Since an eigenspace is a vector subspace, it must contain a set of orthonormal basis vectors.

 

Question

Show that all orthonormal basis eigenvectors in an eigenspace are linearly independent of one another.

Answer

A set of eigenvectors is linearly independent if the only solution to eq1 is when for all . Taking the dot product of eq1 with gives

which reduces to because . Since is arbitrary, we conclude that and that the set is linearly independent.

 

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