### 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#.

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

Invariance of kernel and image under commuting linear mapsAssume 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#.

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#:

Invariance of a spanLet #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#.

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:

Matrix form in case of an invariant subspaceLet #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}#.

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 , 1 , 0 , 0 }# we have #A\,\vec{v}= \rv{0 , 0 , 0 , 1 }#. Thus, the matrix whose columns are the spanning vectors of #U# and the column vector for #A\,\vec{v}# satisfies

\[\text{rank} \matrix{1 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ } = 2 \] Hence, the image #A\,\vec{v}# does belong to #U#.

For #\vec{v}=\rv{0 , 0 , 0 , 1 }# we have #A\,\vec{v}=\rv{-1 , -1 , 0 , 0 }#. Thus, the matrix whose columns are the spanning vectors of #U# and the column vector for #A\,\vec{v}# satisfies

\[\text{rank} \matrix{1 & 0 & -1 \\ 1 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ } = 2 \] Hence, the image #A\,\vec{v}# also belongs to #U#.

We conclude that #U# is invariant under #L_A#. Hence, the answer is Yes.

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