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

### Conjugate matrices

If #\alpha# and #\beta# are bases of a finite dimensional vector space #V#, and the matrix #L_\alpha# of a linear map #L: V\to V# with respect to #\alpha# is given, then, according to the theorem *Basis transition*, the matrix #L_\beta# of #L# with respect to #\beta# can be calculated by use of the formula

\[L_\beta =T L_\alpha T^{-1}\]

where #T = {}_{\beta}I_{\alpha}# is the *transition matrix* of the basis #\alpha# to the basis #\beta#. This means we only have to choose the coordinates once, after that we are able to define all other matrix representations of the linear map #L# by only calculating with matrices. We make this explicit in the following theorem.

Basis transitions in terms of matricesLet #\alpha# be a basis of an #n#-dimensional vector space #V#, where #n# is a natural number, and let #L: V\to V# be a linear map with matrix #A# with respect to #\alpha#.

An #(n\times n)#-matrix #B# is the matrix of #L# with respect to a basis #\beta# for #V# if and only if there exists an invertible #(n\times n)#-matrix #T# with

\[B =T A T^{-1}\]

In particular, the determinant of every matrix determining #L# has the same value, so that we can speak of the **determinant** of the linear map #L#.

We give a special name, conjugate, to matrices #A# and #B# who are both from a given linear map. We will also show that this relation satisfies three important properties, summarized in the notion of equivalence relation:

ConjugateLet #n# be a natural number. Two #(n\times n)#-matrices #A# and #B# are said to be **conjugate** if there exists an invertible #(n\times n)#-matrix #T# with #B = T A\, T^{-1}#. We say that the matrix #T# **conjugates** the matrix #A# to #B# and #T# is called the **conjugator**.

Being conjugate is an **equivalence relation**; this means that it has the following three properties for each three #(n\times n)#-matrices #A#, #B#, #C#:

**Reflexivity**: #A# is conjugate with itself (hence, with #A#)**Symmetry**: If #A# and #B# are conjugate, then #B# and #A# are also conjugate**Transitivity**: If #A# and #B# are conjugate, and #B# and #C# are conjugate, then #A# and #C# are also conjugate.

The matrix #T# is invertible since #\det(T)\ne0#. It is easy to verify that #T# satisfies \(B = T\, A\, T^{-1}\):

\[\begin{array}{rcl} T\, A\, T^{-1} &= & \matrix{-2 & 40 \\ 0 & 17 \\ }\,\matrix{2&-2\\ 2 &1} \, \matrix{-{{1}\over{2}} & {{20}\over{17}} \\ 0 & {{1}\over{17}} \\ }\\

&= & \matrix{76 & 44 \\ 34 & 17 \\ } \, \matrix{-{{1}\over{2}} & {{20}\over{17}} \\ 0 & {{1}\over{17}} \\ }\\ &=& \matrix{-38 & 92 \\ -17 & 41 \\ }\\ &=& B

\end{array}\]

To find #T#, we first solve the matrix equation:

\[\begin{array}{rcl} B &=& T\, A\, T^{-1}\\

&&\phantom{xxx}\color{blue}{\text{the equation that expresses conjugacy}}\\

B \, T&=& T\, A\\

&&\phantom{xxx}\color{blue}{\text{both sides of right hand side multiplied by }T}\\

\matrix{92 z-38 x & 92 w-38 y \\ 41 z-17 x & 41 w-17 y \\ }&=& \matrix{2 y+2 x & y-2 x \\ 2 z+2 w & w-2 z \\ }\\

&&\phantom{xxx}\color{blue}{\text{matrices multiplied, where }T=\matrix{x&y\\ z&w}}\\

\end{array}\] The solution of this system of linear equations in #x#, #y#, #z#, and #w# is

\[

x={{39 z-2 w}\over{17}} ,\quad y={{2 z+40 w}\over{17}}

\] Hence, we have the free parameters # w , z #. If we choose # w=17 , z=0 #, then we find # x=-2 , y=40 #, so

\[ T = \matrix{x&y\\ z&w} = \matrix{-2 & 40 \\ 0 & 17 \\ }\] This answer works because the matrix is invertible.

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.