### Invariant subspaces of linear maps: Matrices and coordinate transformations

### Matrix equivalence

*Conjugation* is about matrices representing a linear map from a linear vector space to the same vector space. Here, we discuss the case of two different vector spaces or, more generally, two different bases: one for the domain #V# and one for the range #W# of a linear map #V\to W#. We will see that the problem of determining whether two matrices represent the same linear map with respect to suitable bases for the domain and range, is much easier than the case where the same basis must be used for both domain and range.

Matrix equivalence Let #m# and #n# be natural numbers. Two #(m\times n)#-matrices #A# and #B# are called **matrix equivalent** if there is an invertible #(m\times m)#-matrix #S# and an invertible #(n\times n)#-matrix #T# such that #B = S\, A\, T#.

- Two matrices #A# and #B# of the same size are matrix equivalent if and only if they have the same rank. In particular, matrix equivalence is an
*equivalence relation*. - Let #\alpha# be a basis for an #n#-dimensional vector space #V#, let #\beta # be a basis for an #m#-dimensional vector space #W#, and let #L: V\to W# be a linear map having matrix #A# with respect to #\alpha# and #\beta#. An #(m\times n)#-matrix #B# is the matrix of #L# with respect to a basis for #V# and a basis for #W# if and only if #A# and #B# are matrix equivalent.

We give some other characterizations of linear maps with equal rank. Recall from the theory *The matrix of a linear map* that, for a linear map #L:V\to W# and bases #\alpha# for #V# and #\beta# for #W#, the matrix of #L# with respect to #\alpha# and #\beta# is denoted by #{}_\beta L_\alpha#, and, in case #V=W# and #\alpha = \beta#, also by #L_\alpha#.

Criteria for matrix equivalence Let #V# and #W# be vector spaces of finite dimension #n# and #m#, respectively, and let #L# and #M# be linear maps #V\to W#. The following statements are equivalent.

- There are isomorphisms #P: V\to V# and #Q:\ W\to W# such that #M = Q\,L\,P#.
- There are bases #\beta_1# and #\beta_2# for #V# and #\gamma_1# and #\gamma_2# for #W# such that #{}_{\gamma_1}L_{\beta_1} = {}_{\gamma_2}M_{\beta_2}#.
- There are bases #\beta# for #V# and #\gamma# for #W#, such that #{}_{\gamma}L_{\beta} # and # {}_{\gamma}M_{\beta}# are matrix equivalent.
- For each pair of bases #\beta# for #V# and #\gamma# for #W# the matrices #{}_{\gamma}L_{\beta} # and # {}_{\gamma}M_{\beta}# are matrix equivalent.
- For each pair of bases #\beta# for #V# and #\gamma# for #W# the matrices #{}_{\gamma}L_{\beta} # and # {}_{\gamma}M_{\beta}# have the same rank.

According to the theorem

*Matrix equivalence*, the matrices are matrix equivalent if and only if they have the same rank. The rank of #A# is #0 # and the rank of #B# is #2#. Therefore, the answer is No.

After all,

\[\begin{array}{rcl}B &=& \matrix{-1 & -2 \\ 1 & 3 \\ }\, \matrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ }\, \matrix{3 & 1 & -5 \\ -2 & -1 & 3 \\ 1 & 0 & -1 \\ }\end{array}\]

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