Conjugate matrices

Matrix calculus: 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.

Let $\alpha$ 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 $L$ .

Assume that $\beta$ is a basis for $V$ , so $B=L_\beta$ , that is, $B$ is the matrix of $L$ with respect to $\beta$ . Then theorem Basis transition tells us that $A = L_\alpha$ and $B = L_\beta$ are related by means of $B =\left({}_\beta I_\alpha \right) A \left({}_\beta I_{\alpha}\right)^{-1}$ . In this case $T = {}_\beta I_\alpha$ is an invertible $(n\times n)$ -matrix with $B= T A T^{-1}$ .

The other way around: assume that we have an invertible $(n\times n)$ -matrix $T$ with $B= T A T^{-1}$ . Now choose $\beta = L_T \alpha$ , such that $T = \left(\beta\alpha^{-1}\right)_{\varepsilon}$ . Then $\basis{\beta^{-1}(\varepsilon_1),\ldots,\beta^{-1}(\varepsilon_n)}$ is the corresponding basis of $V$ (because $\beta$ is invertible with inverse $\alpha^{-1}L_{T^{-1}}$ ), and it follows that

$L_\beta = {}_\beta I_\alpha\, A\, \left({}_\beta I_\alpha\right)^{-1} = \left(\beta\alpha^{-1}\right)_\varepsilon A \left(\beta\alpha^{-1}\right)_\varepsilon^{-1} = T AT^{-1} = B$

We have shown that $B$ is the matrix of $L$ with respect to the basis $\beta$ for $V$ .

To prove the last statement, we show that $B$ and $A$ have the same determinant. By using the properties of the determinant established earlier we find

$\det(B) =\det\left(T AT^{-1} \right) =\det(T)\cdot\det( A)\cdot\det(T^{-1}) =\det( A)\cdot\det(T)\cdot\det(T)^{-1}=\det(A)$

A consequence of this theorem is the known fact that the $(n\times n)$ -zero matrix is the only matrix of the zero map of an $n$ -dimensional vector space to itself.

Another consequence of this theorem is the known fact that the $(n\times n)$ -identity matrix is the only matrix of the identity map of an $n$ -dimensional vector space to itself.

Yet another consequence is the fact that for each linear map of a $1$ -dimensional vector space to itself we have a unique matrix. After all, that map has to be a scalar multiplication and the matrix has to be the scalar by which the vector is multiplied.

Determinant If two linear maps $V\to V$ have different determinants, then it is not possible to find two bases of $V$ , such that the matrix of the one linear map with respect to the first basis is equal to the matrix of the other linear map with respect to the second basis.

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:

Let $n$ 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.

Reflexivity: For $T$ take the $(n\times n)$ -matrix identity matrix $I$ . Then we have $A = I A I^{-1}$ . Hence, $A$ and $A$ are conjugate.

Symmetry: Assume that $A$ and $B$ are conjugate. Then an invertible $(n\times n)$ -matrix $T$ with $B= TAT^{-1}$ exists. By multiplying both sides on the left by $T^{-1}$ and on the right by $T$ , we see that $T^{-1} B T = A$ , or

$A = \left(T^{-1}\right) B\left(T^{-1}\right)^{-1}$ Since $T^{-1}$ is an invertible $(n\times n)$ -matrix, we conclude that $B$ and $A$ are conjugate.

Associativity: Assume that $A$ and $B$ are conjugate, and that $B$ and $C$ are also conjugate. Then we have invertible $(n\times n)$ -matrices $S$ and $T$ , such that $B = SAS^{-1}$ and $C = TBT^{-1}$ . Consequently the invertible $(n\times n)$ -matrix $T\,S$ satisfies

$C = TBT^{-1} = TSAS^{-1}T^{-1} = \left(TS\right)A\left(T S\right)^{-1}$ such that $A$ and $C$ are conjugate.

Example

Determining if two $(n\times n)$ -matrices $A$ and $B$ are conjugate largely boils down to solving linear equations: we start by solving the matrix equation

$BT=TA\phantom{xxx}\text{in the unknown }\phantom{xxx}T$ and next look for an invertible matrix among the solutions. Here is an example: Assume

$A = \matrix{1&0\\ 0& -1}\phantom{xxx}\text { and }\phantom{xx} B = \matrix{1&1\\ 0& 1}$ If $A$ and $B$ are conjugate, then there exists an invertible $(2\times2)$ -matrix $T$ with $B = TAT^{-1}$ .

After multiplying from the right by $T$ we have the linear matrix equation

$BT=TA$ If we write

$T= \matrix{x&y\\ z&w}$ then, after performing the matrix multiplications, the matrix equation transforms in

$\matrix{x+z& y+w\\ z & w} = \matrix{x&-y\\ z&-w}$ and hence, in the system of linear equations

$\lineqs{ x+z &=& x\\ y+w &=& -y\\ z &=& z\\ w &=& -w}$ This system has the solution

$T = \matrix{x&y\\ z&w} = \matrix{x&0\\ 0&0}$ where $x$ is a free parameter. The matrix $T$ is not invertible, so $A$ and $B$ are not conjugate. In other words: There is no basis of $\mathbb{R}^2$ with respect to which the linear map with matrix $A$ with respect to the standard basis has the matrix $B$ .

In this case you can see directly that $A$ and $B$ are not conjugate because their determinants differ. Later we discuss the determination of the Jordan normal form of a square matrix; this method is much more efficient.

With given $(n\times n)$ -matrices $A$ and $B$ and invertible $(n\times n)$ -matrix $T$ such that $B=TAT^{-1}$ , the conjugator $T$ is not unique: any scalar multiple of $T$ (that is not equal to the zero matrix) is also a conjugator from $A$ to $B$ . Furthermore, it may happen that various matrices that are not scalar multiples of each other, are all conjugators from $A$ to $B$ . For example, take $n=2$ and $A=B=I_2$ . Then every invertible $(2\times 2)$ -matrix satisfies $B=TAT^{-1}$ by definition of the inverse matrix, because $I_2= TT^{-1}=TI_2T^{-1}$ .

Similar Another commonly used name for conjugate matrices is similar matrices.

The set $M_{n\times n}$ of $(n\times n)$ -matrices can be divided in separate disjoint parts, each consisting of matrices that are mutually conjugate and not conjugate to any matrix of another part. This is a property that holds for every equivalence relation. In general, these parts are called equivalence classes and usually are also named after the specific equivalence relation, hence, here conjugacy classes.

Let $V$ be a vector space of finite dimension $n$ . For each linear map $L:V\to V$ the matrices in $M_{n\times n}$ that determine $L$ with respect to a suitable basis, form a conjugacy class.

The conjugacy class of the identity map consists of only $I_n$ . The only linear maps of which the conjugacy class consists of a single matrix, are scalar multiplications. After all, the matrix corresponding to a scalar multiplication by $\lambda$ is the diagonal matrix $D_\lambda=\lambda\cdot I_n$ of which all diagonal elements are equal to $\lambda$ . This matrix is the only one that commutates with all other $(n\times n)$ -matrices. Therefore $D_\lambda$ only conjugates with itself: $TD_\lambda T^{-1}=D_\lambda$ for each invertible matrix $T$ .

Determinant From the theorem above and properties of the determinant it follows that two matrices with different determinants are not conjugate. This can also be seen directly: if $A$ and $B$ are conjugate $(n\times n)$ -matrices, then there exists an invertible $(n\times n)$ -matrix $T$ such that $A = T\, B\, T^{-1}$ . From known properties of the determinant it then follows that

$\det(A) = \det(T\, B\, T^{-1}) = \det(T)\cdot\det( B)\cdot\det( T^{-1}) = \det(T)\cdot\det( B)\cdot\det( T)^{-1} = \det( B)$

Invariants Since conjugate matrices represent the same linear map, they share all properties of that map. We saw above that conjugate matrices have the same determinant. Other properties that conjugate matrices share, are trace, characteristic polynomial, rank and minimal polynomial. Two matrices of which one or more of these properties differ, are not conjugate.

A property of a matrix of a linear map that does not depend on the chosen basis is called an invariant.

Thanks to a statement we will discuss later, the characteristic polynomial directly shows whether or not two $(2\times 2)$ -matrices are conjugate. However, that is not the case for matrices of larger dimensions.

The following two matrices $A$ and $B$ are conjugate:

$A = \matrix{1&2\\ 3 &4}\phantom{xxx}\text{ and }\phantom{xxx} B = \matrix{43 & 68 \\ -24 & -38 \\ }$ Determine a conjugator $T$ from $A$ to $B$ ; that is, an invertible $(2\times2)$ -matrix $T$ such that

$B = T\, A\, T^{-1}$

$T =$ $\matrix{-3 & -42 \\ 0 & 24 \\ }$

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

$\begin{array}{rcl} T\, A\, T^{-1} &= & \matrix{-3 & -42 \\ 0 & 24 \\ }\,\matrix{1&2\\ 3 &4} \, \matrix{-{{1}\over{3}} & -{{7}\over{12}} \\ 0 & {{1}\over{24}} \\ }\\ &= & \matrix{-129 & -174 \\ 72 & 96 \\ } \, \matrix{-{{1}\over{3}} & -{{7}\over{12}} \\ 0 & {{1}\over{24}} \\ }\\ &=& \matrix{43 & 68 \\ -24 & -38 \\ }\\ &=& B \end{array}$
To find $T$ , we first solve the matrix equation:

$\begin{array}{rcl} B &=& T\, A\, T^{-1}\\ &&\phantom{xxx}\color{blue}{\text{the equation that expresses conjugacy}}\\ B \, T&=& T\, A\\ &&\phantom{xxx}\color{blue}{\text{both sides of right hand side multiplied by }T}\\ \matrix{68 z+43 x & 43 y+68 w \\ -38 z-24 x & -24 y-38 w \\ }&=& \matrix{3 y+x & 4 y+2 x \\ z+3 w & 2 z+4 w \\ }\\ &&\phantom{xxx}\color{blue}{\text{matrices multiplied, where }T=\matrix{x&y\\ z&w}}\\ \end{array}$ The solution of this system of linear equations in $x$ , $y$ , $z$ , and $w$ is

$x={{-13 z-w}\over{8}} ,\quad y={{-z-21 w}\over{12}}$ Hence, we have the free parameters $w , z$ . If we choose $w=24 , z=0$ , then we find $x=-3 , y=-42$ , so

$T = \matrix{x&y\\ z&w} = \matrix{-3 & -42 \\ 0 & 24 \\ }$ This answer works because the matrix is invertible.

The answer is not unique. Other choices than $w=24 , z=0$ for the entries of $T=\matrix{x&y\\ z&w}$ are possible.

New example