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 $W=\{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}},\boldsymbol{\mathit{v_3}}\}$ of the vector space $V=\left \{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}}, \cdots,\boldsymbol{\mathit{v_k}}\right \}$ 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 $W$ is another vector in $W$ .

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

For example, in the $\mathbb{R}^{3}$ space, a plane through the origin is a subspace, as is a line through the origin. Hence, $\mathbb{R}^{2}$ and $\mathbb{R}$ are subspaces of $\mathbb{R}^{3}$ . The entire $\mathbb{R}^{3}$ space and the single point at the origin are also subspaces of the $\mathbb{R}^{3}$ 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 $Ax_i=\lambda_ix_i$ , where $i=1,2,3$ . If the eigenvalues of $x_1$ , $x_2$ and $x_3$ are $\lambda_1=1$ , $\lambda_2=2$ and $\lambda_3=2$ respectively, then $J=\{x_1\}$ and $K=\{x_2,x_3\}$ are the eigenspaces of the operator $A$ . Since an eigenspace is a vector subspace, it must contain a set of orthonormal basis vectors.

Question

Show that all orthonormal basis eigenvectors $\boldsymbol{\mathit{v_{k}}}$ in an eigenspace $S=\{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}},\cdots,\boldsymbol{\mathit{v_k}}\}$ are linearly independent of one another.

Answer

A set of eigenvectors is linearly independent if the only solution to eq1 is when $c_k=0$ for all $k$ . Taking the dot product of eq1 with $\boldsymbol{\mathit{v_{i}}}$ gives

$c_1\boldsymbol{\mathit{v_{1}}}\cdot\boldsymbol{\mathit{v_{i}}}+c_2\boldsymbol{\mathit{v_{2}}}\cdot\boldsymbol{\mathit{v_{i}}}+\cdots +c_i\boldsymbol{\mathit{v_{i}}}\cdot\boldsymbol{\mathit{v_{i}}}+\cdots +c_k\boldsymbol{\mathit{v_{k}}}\cdot\boldsymbol{\mathit{v_{i}}}=0$

which reduces to $c_i=0$ because $\boldsymbol{\mathit{v_{i}}}\cdot\boldsymbol{\mathit{v_{i}}}\neq 0$ . Since $i$ is arbitrary, we conclude that $c_1=c_2=\cdots =c_k=0$ and that the set $\{\boldsymbol{\mathit{v_{1}}},\boldsymbol{\mathit{v_{2}}},\cdots,\boldsymbol{\mathit{v_k}}\}$ is linearly independent.