The notion of invariant subspace

Invariant subspaces of linear maps: Invariant subspaces

The notion of invariant subspace

The line through the origin spanned by the eigenvector of a linear map is transformed into itself. Such subspaces are called invariant. If there are no or insufficient eigenvectors, there might still be invariant subspaces of higher dimension. For example, think of a rotation around a line through the origin in $\mathbb{R}^3$ , in which case the plane through the origin perpendicular to the rotation axis is such an invariant subspace.

These kind of invariant subspaces also lead to simple matrix representations.

Invariant subset

Let $X$ be a subset of a set $V$ and $G$ a map $V\rightarrow V$ .

Then $X$ is called invariant under $G$ if $G( \vec{x})\in X$ for every $\vec{x}\in X$ .

In this case all images $G(\vec{x})$ with $\vec{x}\in X$ again lie in $X$ . Hence, there exists a map $X\rightarrow X$ with the same mapping rule as $G$ . This map is called the restriction of $G$ to $X$ . We denote this restriction by $\left. G\right|_X$ .

Function examples Let $f:\mathbb{R}\to\mathbb{R}$ be the map with function rule $f(x) = x^2$ . Then the set of non-negative numbers is a subset of $\mathbb{R}$ which is invariant under $f$ . Also the singleton $\{0\}$ and the intervals $\ivco{1}{\infty}$ and $\ivco{0}{1}$ are invariant subspaces.

Let $V$ be a vector space and $L:V\to V$ a linear map.

The nullspace $\ker{L}$ is invariant: if $\vec{x}\in\ker{L}$ , then $L(\vec{x})=\vec{0}$ and this vector is in $\ker{L}$ .
The image $\im{L}$ is also an invariant subspace: if $\vec{y}\in\im{L}$ , then $L( \vec{y})$ is also in $\im{L}$ .
If $\lambda$ is an eigenvalue of $L$ , then the corresponding eigenspace $E_\lambda$ is invariant under $L$ : if $\vec{z}\in E_\lambda$ , then $L(\vec{z})=\lambda \vec{z}\in E_\lambda$ since $E_\lambda$ is a subspace of $V$ .

The following observation enables us to find a lot of invariant subspaces under linear maps.

Assume that $L$ and $M$ are both linear maps $V\to V$ that commute with each other (meaning that $L\, M = M\,L$ ). Then $\ker{M}$ and $\im{M}$ are invariant subspaces under $L$ .

$\ker{M}$ : if $\vec{x}\in\ker{M}$ , then $M (L(\vec{x})) =M\, L(\vec{x}) =L\,M(\vec{x})=L(M(\vec{x})) =L(\vec{0}) =\vec{0}$ so $L(\vec{x})$ belongs to $\ker{M}$ . This proves that $\ker{M}$ is invariant under $L$ .

$\im{M}$ : if $\vec{y}\in\im{M}$ , then there exists a vector $\vec{z}$ such that $\vec{y} = M(\vec{z})$ . Now we have $L(\vec{y}) =L(M(\vec{z}))=L\,M(\vec{z})=M\, L(\vec{z}) =M(L(\vec{z}))$ so $L(\vec{y})$ belongs to $\im{M}$ . This proves that $\im{M}$ is invariant under $L$ .

We will apply this result with $M= p(L)$ for a polynomial $p(x)$ that divides the characteristic polynomial $p_L(x)$ of $L$ . The case $p(x) = (\lambda-x)^k$ , where $\lambda$ is an eigenvalue of $L$ and $k$ a natural number less than or equal to the multiplicity of the root $\lambda$ of $p_L(x)$ , is an important example.

The case $L=M$ is already dealt with as an example in the above definition of invariant subset under the tab Linear examples.

The third example, of the eigenspace $E_\lambda$ of $L$ for an eigenvalue $\lambda$ , is a special case, with $M=L-\lambda\,I_V$ , because $E_\lambda=\ker{M}$ .

If $X$ is a linear subspace and $L$ a linear map, then you do not need to check that all vectors of $X$ under $L$ end up in $X$ in order to determine that $X$ is invariant under $L$ :

Let $L:V\rightarrow V$ be a linear map and suppose $W=\linspan{\vec{a}_1,\ldots,\vec{a}_n}$ . The subspace $W$ is invariant under $L$ if and only if $L( \vec{a}_i)\in W$ for $i=1,\ldots ,n$ .

If $W$ is invariant, then $L( \vec{w})\in W$ for all $\vec{w}\in W$ . Choosing $\vec{w}=\vec{a}_i$ , we find that $L(\vec{a}_i)\in\linspan{\vec{a}_1,\ldots ,\vec{a}_n}$ for all $i=1,\ldots,n$ .

Conversely, assume that $L( \vec{a}_i)\in W$ for $i=1,\ldots,n$ and let $\vec{w}\in W$ be arbitrary. Then $\vec{w}$ can be written as $\vec{w}=w_1\vec{a}_1+\cdots +w_n\vec{a}_n$ , so $L( \vec{w})=w_1 L( \vec{a}_1) +\cdots +w_n L( \vec{a}_n)$ . Since all vectors $L( \vec{a}_i)$ lie in $W$ , it follows that $L( \vec{w})\in W$ .

It follows from the theorem that, if $\basis{\vec{a}_1,\ldots,\vec{a}_n}$ is a basis of the subspace $W$ , the subspace $W$ is invariant under $L$ if and only if the rank of the matrix with columns $\vec{a}_1,\ldots,\vec{a}_n,L(\vec{a}_1),\ldots,L(\vec{a}_n)$ equals $n$ . This gives an efficient method for determining whether or not a subspace is invariant under a linear map.

By choosing a matrix of a linear map with respect to a basis which partly consists of a basis for an invariant subspace, we can make sure that a number of matrix elements are equal to zero:

Let $L :V\rightarrow V$ be linear and assume that $\alpha=\basis{\vec{a}_1,\ldots ,\vec{a}_n}$ is a basis for $V$ such that $W=\linspan{\vec{a}_1,\ldots ,\vec{a}_m}$ is invariant under $L$ . Then the matrix $L_\alpha$ is of the form $L_\alpha = \matrix{B&*\\ 0 &*}$ where $0$ indicates a submatrix with all zeros, each $*$ an arbitrary matrix and $B$ the $(m\times m)$ -matrix of ${\left.L\right|_W}:W\rightarrow W$ with respect to the basis $\basis{\vec{a}_1,\ldots ,\vec{a}_m}$ .

According to the previous theorem, we have, for $i=1,\ldots,m$ , $L( \vec{a}_i)=a_{i1}\vec{a}_1+\cdots +a_{im}\vec{a}_m+0\vec{a}_{m+1}+\cdots +0\vec{a}_n$ From this, the theorem follows directly.

Determine whether the linear subspace $U= \linspan{ \rv{1 , 0 , 0 , 0 },\rv{-1 , 0 , 1 , 1 }}$ of $\mathbb{R}^4$ is invariant under the linear map $A : \mathbb{R}^4\to\mathbb{R}^4$ determined by the matrix $A = \matrix{-1 & 5 & -1 & -1 \\ 0 & 1 & 0 & 0 \\ 1 & -3 & 0 & 3 \\ 1 & -3 & 1 & 2 \\ }$

Yes

The linear subspace $U$ is invariant under $L_A$ if and only if each of the images of the spanning vectors of $U$ belongs to $U$ .

To determine if the image under $L_A$ of a vector $\vec{v}$ belongs to $U$ , we will check if the rank of the matrix whose columns are the spanning vectors of $U$ remains constant ( $2$ ) if we add the column vector $A\,\vec{v}$ .

For $\vec{v}=\rv{1 , 0 , 0 , 0 }$ we have $A\,\vec{v}= \rv{-1 , 0 , 1 , 1 }$ . Thus, the matrix whose columns are the spanning vectors of $U$ and the column vector for $A\,\vec{v}$ satisfies
$\text{rank} \matrix{1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ } = 2$ Hence, the image $A\,\vec{v}$ does belong to $U$ .
For $\vec{v}=\rv{-1 , 0 , 1 , 1 }$ we have $A\,\vec{v}=\rv{-1 , 0 , 2 , 2 }$ . Thus, the matrix whose columns are the spanning vectors of $U$ and the column vector for $A\,\vec{v}$ satisfies
$\text{rank} \matrix{1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \\ } = 2$ Hence, the image $A\,\vec{v}$ also belongs to $U$ .
We conclude that $U$ is invariant under $L_A$ . Hence, the answer is Yes.

New example