From real to complex vector spaces and back

Invariant subspaces of linear maps: Invariant subspaces

From real to complex vector spaces and back

Before we will deal with normal forms for real matrices having a characteristic polynomial with non-real roots, we will discuss the extension of a real vector space towards a complex one and the question how the original real vector space can be found in the complex one.

Let $V$ be a real vector space. Take a look at the set

$V_{\mathbb{C} }= V + \ii V$

consisting of all vectors of the form $\vec{x}+\ii\vec{y}$ , where $\vec{x}$ and $\vec{y}$ are vectors of $V$ , and the addition is formal (hence, can also be seen as the pair $\rv{\vec{x},\vec{y}}$ ). This is a complex vector space if it is provided with the following operations:

$\begin{array}{rclcl}\text{scalar multiplication}&:&(a+b\ii)(\vec{x}+\ii\vec{y}) &=&(a\cdot\vec{x}-b\cdot\vec{y})+\ii(b\cdot\vec{x}+a\cdot\vec{y})\\ \text{vector addition}&:&(\vec{x}+\ii\vec{y}) +(\vec{u}+\ii\vec{v})&=&(\vec{x}+\vec{u})+\ii(\vec{y}+\vec{v})\end{array}$ We will call this the extension of $V$ to a complex vector space.

The map $\sigma : V_{\mathbb{C} }\to V_{\mathbb{C} }$ defined by $\sigma(\vec{x}+\ii\vec{y}) = \vec{x}-\ii\vec{y}$ for $\vec{x},\vec{y}$ in $V$ is called complex conjugate. This map is semi-linear, which means that, for all complex numbers $\lambda$ and $\mu$ and all vectors $\vec{u}$ and $\vec{w}$ of $V_{\mathbb{C} }$ we have

$\sigma(\lambda\cdot\vec{u}+\mu\cdot\vec{w}) = \overline{\lambda}\cdot\sigma(\vec{u}) +\overline{\mu}\cdot\sigma(\vec{w})$ where $\overline{\lambda}$ indicates the complex conjugate of $\lambda$ .

If $\alpha$ is a basis for $V$ , then $\alpha$ is also a basis for $V_{\mathbb{C} }$ . In particular $V$ and $V_{\mathbb{C} }$ have the same dimension.

If $A$ is a real $(n\times n)$ -matrix, then $L_A$ can be interpreted as a linear map $V\to V$ , where $V=\mathbb{R}^n$ or $V = \mathbb{C}^n$ . In the case $V=\mathbb{R}^n$ the map $\left(L_A\right)_{\mathbb C}$ is equal to the map $L_A: \mathbb{C}^n\to \mathbb{C}^n$ of the second case. There is little confusion between the interpretation of $L_A$ as a linear map with a real vector space as domain and as a linear map on a complex vector space.

The fact that $\sigma$ is semi-linear follows from the following two calculations, where we will write $\Re(\vec{x}+\ii\vec{y})=\vec{x}$ and $\Im(\vec{x}+\ii\vec{y}) =\vec{y}$ for $\vec{x},\vec{y}$ in $V$ :

$\begin{array}{rcl} \sigma(\vec{u}+\vec{w}) &=& \sigma( \Re(\vec{u})+\ii \Im(\vec{u})+\Re(\vec{w})+\ii\Im(\vec{w}) )\\&=& \sigma( \Re(\vec{u})+\Re(\vec{w})+\ii(\Im(\vec{u})+\Im(\vec{w}) ))\\&=&\Re(\vec{u})+\Re(\vec{w})-\ii(\Im(\vec{u})+\Im(\vec{w}) )\\&=&\Re(\vec{u})-\ii\Im(\vec{u})+\Re(\vec{w})-\ii\Im(\vec{w}) \\ &=& \sigma(\vec{u}) +\sigma(\vec{w})\\ \\ \sigma(\lambda\cdot\vec{u}) &=& \sigma( (\Re\lambda+\ii\Im\lambda)(\Re(\vec{u})+\ii \Im(\vec{u})) )\\&=& \sigma( \Re\lambda\cdot\Re(\vec{u}) -\Im\lambda\cdot\Im(\vec{u})+\ii (\Im\lambda\cdot\Re(\vec{u}) +\Re\lambda\cdot \Im(\vec{u})))\\ &=&\Re\lambda\cdot\Re(\vec{u}) -\Im\lambda\cdot\Im(\vec{u})-\ii (\Im\lambda\cdot\Re(\vec{u}) +\Re\lambda\cdot \Im(\vec{u}))\\&=&(\Re\lambda-\ii\Im\lambda)\cdot(\Re(\vec{u}) -\ii \Im(\vec{u}))\\ &=& \overline{\lambda}\cdot \sigma(\vec{u}) \end{array}$

With the term complex conjugate it's possible to get confused: complex conjugation of the elements of the matrix $A = \matrix{2+3\ii&0\\ 0&2-3\ii}$ gives $\sigma(A) = \matrix{2-3\ii&0\\ 0&2+3\ii}$ while conjugation over $\mathbb{C}$ of $A$ results in a matrix of the form $T\,A\,T^{-1}$ where $T$ is an invertible $(2\times2)$ -matrix. The second operation is explicitly not named 'complex conjugate'. In the special case where $T$ is a permutation matrix corresponding to the permutation $[1,2]$ the two definitions are the same.

To view a complex vector space as a real space, no expansion is needed:

A complex vector space $W$ can be seen as a real vector space by limitation of scalars. This means we will only view the real numbers of $W$ as scalars. We will write $\left.W\right|_{\mathbb{R}}$ to indicate this real vector space.

Let $V$ be a real vector space. After limitation of scalars for $W = V_{\mathbb{C} }$ the semi-linear map complex conjugate $\sigma:W\to W$ becomes a linear map $\left.W\right|_{\mathbb{R}}\to\left.W\right|_{\mathbb{R}}$ with the property that $V$ is the eigenspace of $\sigma$ at eigenvalue $1$ and $\ii V$ the eigenspace of $\sigma$ at eigenvalue $-1$ .

If $\alpha$ is a basis for $W$ , then $\alpha\cup\ii\alpha$ is a basis for $\left.W\right|_{\mathbb{R}}$ . In particular, the dimension of $\left.W\right|_{\mathbb{R}}$ is twice as large as the dimension of $W$ .

If we start with a real vector space $V$ , extend the scalars to obtain $V_{\mathbb{C} }$ and next restrict the scalars, we get the real vector space $U = \left.V_{\mathbb{C} }\right|_{\mathbb{R} }$ of which $V$ is indeed a subspace, but of dimension twice the dimension of $V$ . The question how to recover $V$ from $U$ can be answered in multiple ways. The statement about $\sigma$ gives one method: the minimal polynomial of $\sigma$ is equal to $x^2-1$ , so

$U = \ker{C-I_U}\oplus\ker{C+I_U},\phantom{xxx}\text{ where }\phantom{xxx} V =\ker{C-I_U}$

The other term of the direct sum is $\ker{C+I_U} = \ii V$ .

If $L:V\to V$ , then $L$ determines a unique linear map $L_{\mathbb{C}}:V_{\mathbb{C} }\to V_{\mathbb{C} }$ . This map commutes with $C$ , so that $V = \ker{C-I_U}$ is invariant under $L_{\mathbb{C}}$ . Furthermore, the limitation of $L_{\mathbb{C}}$ to $V$ is equal to $L$ . Since the characteristic polynomial of $L_{\mathbb{C}}$ is a product of linear factors, we can find a basis $\alpha$ for $V_{\mathbb{C} }$ , in such a way that the matrix $\left(L_{\mathbb{C}}\right)_\alpha$ has a Jordan form. With help of the real and imaginary parts of this basis, we will find a basis $\beta$ of $V$ so that $L_\beta = \left(\left.L_{\mathbb{C}}\right|_V\right)_{\beta}$ has a form closely resembling the Jordan normal form.

Take a look at the diagonal matrix $A = \matrix{\lambda&0\\ 0&\overline{\lambda}}$ where $\lambda$ is a non-real complex number and $\overline\lambda$ the complex conjugate of $\lambda$ . This matrix is conjugate with the real matrix $B = \matrix{\Re\lambda & -\Im\lambda\\ \Im\lambda & \Re\lambda}$ This follow directly from the fact that the characteristic polynomials of $A$ and $B$ are both equal to $x^2-2\Re\lambda \cdot x+\lambda\cdot\overline\lambda$ , because two $(2\times2)$ -matrices with two corresponding (but distinct) eigenvalues are always conjugate.

If we start with the matrix $B$ , we will find the diagonal form $A$ after determining the complex eigenvalues.

Vice versa, we can, starting with $A$ as above, find a real matrix conjugate to $A$ (such as $B$ ) in the following manner: we search for a 2-dimensional real subspace $V$ of $\left.{\mathbb{C}^2}\right|_{\mathbb{R}}$ which is kept invariant by $A$ . This boils down to determining an invertible $(2\times2)$ -matrix $T=(t_{ij})$ such that $V$ is spanned by the columns vectors $T(\vec{e}_1)$ and $T(\vec{e}_2)$ . The matrix of $L_A$ relative to the basis $\basis{\matrix{t_{11}\\ t_{21}},\matrix{t_{12}\\ t_{22}}}$ for $V$ must have real elements. This matrix is $T^{-1}AT$ , hence, the complex conjugate $\sigma(T^{-1}AT)$ of this matrix must be equal to $T^{-1}AT$ . Furthermore, the complex conjugate $\sigma(A)$ of $A$ is conjugated to $A$ through $P=\matrix{0&1\\ 1&0}$ . We will use this to find a solution for $T$ :

$\begin{array}{rcl}\sigma(T^{-1}AT)&=&T^{-1}AT\\ &&\phantom{xxx}\color{blue}{\text{this matrix must have real elements }}\\ \sigma(T^{-1})\sigma(A)\sigma(T)&=&T^{-1}AT\\ &&\phantom{xxx}\color{blue}{\text{complex conjugate of sum and product is }}\\&&\phantom{xxx}\color{blue}{\text{sum and product of complex conjugate}}\\ \sigma(T^{-1})PAP^{-1}\sigma(T)&=&T^{-1}AT\\ &&\phantom{xxx}\color{blue}{\sigma(A) = PA P^{-1}}\\ \left(T\sigma(T^{-1})P\right)A&=&A\left(T\sigma(T^{-1})P\right)\\&&\phantom{xxx}\color{blue}{\text{ on the left-hand side multiplied by }T\text{ }}\\ &&\phantom{xxx}\color{blue}{\text{ on the right-hand side multiplied by }\sigma(T)^{-1}P=\sigma(T^{-1})P}\\ T\sigma(T^{-1})P&=&I_2\\&&\phantom{xxx}\color{blue}{\text{this is sufficient for a solution (not essential)}}\\ T &=&P\sigma(T)\\&&\phantom{xxx}\color{blue}{\text{ on the right-hand side multiplied by }P^{-1}\sigma(T)\text{ ; }P^{-1} = P}\\\matrix{t_{11}&t_{12}\\ t_{21}& t_{22}} &=&\matrix{\overline{t_{21}}&\overline{t_{22}}\\ \overline{t_{11}}& \overline{t_{12}}}\\&&\phantom{xxx}\color{blue}{T=(t_{ij})\text{ substituted and worked out matrix multiplication }}\\T &=& \matrix{t_{11}&{t_{12}}\\\overline{t_{11}}& \overline{t_{12}}}\\&&\phantom{xxx}\color{blue}{\text{ solution for }t_{21},\ t_{22}\text{ substituted }}\\ T &=& \matrix{1&\ii\\ 1&-\ii}\\&&\phantom{xxx}\color{blue}{\text{ solution }t_{11}=1,\ t_{12}=\ii\text{ chosen such that }T^{-1}\text{ exists}}\\T^{-1}A\,T &=&\dfrac{1}{2\ii} \matrix{\ii&\ii\\ 1&-1}\matrix{\lambda&0\\ 0&\overline{\lambda}} \matrix{1&\ii\\ 1&-\ii}\\ &=&\dfrac{1}{2}\matrix{\lambda+\overline{\lambda}&\ii({\lambda}-\overline\lambda)\\ \ii(\overline{\lambda}-\lambda)&\lambda+\overline{\lambda}} \\&=&\matrix{\Re\lambda&-\Im\lambda\\\Im\lambda&\Re\lambda}\\ &=& B\end{array}$ With this we have found the conjugator $T$ that conjugates $A$ to $B$ .

As we will see later, it is possible to find a unique matrix in every conjugation class of real $(n\times n)$ -matrices, starting with the complex Jordan normal form. However, with the theorem below, it is easy to find out when two real $(n\times n)$ -matrices are conjugate, which is the case when they are conjugate as complex matrices.

If two real $(n\times n)$ -matrices are conjugate over $\mathbb{C}$ , then they are also conjugate over $\mathbb{R}$ .

Assume that two $(n\times n)$ -matrices $A$ and $B$ are conjugate over $\mathbb{C}$ . Then there exists a complex invertible matrix $T$ so that $B = T^{-1}\,A\,T$ . After multiplying from the left by $T$ on both sides we have $A\,T =T\,B$ Write $T = K +\ii \cdot L$ , where $K=\Re T$ and $L = \Im T$ are the real and imaginary part of $T$ . We then have $A\,K = K\,B\phantom{xxx}\text{ and }\phantom{xxx}A\,L = L\,B$ and thus

$A\,(K+x\cdot L) =(K+x\cdot L) \,B\phantom{xxx}\text{ for each number }\phantom{xx}x$

Hence, it is sufficient to derive the existence of a real number $x$ where $K+x\cdot L$ is invertible. Therefore, consider the determinant $\det(K+x\cdot L)$ as a function of $x$ . This is a polynomial function of degree $n$ in $x$ with real coefficients because $K$ and $L$ are real. Since $K+\ii\cdot L=T$ is invertible, $\det(K+\ii\cdot L)\ne0$ so the polynomial function is not the zero function. The polynomial function has no more than $n$ zeros, so that there must exist a real value of $x$ for which $\det(K+x\cdot L)\ne0$ , so that $K+x\cdot L$ is a real invertible matrix. This proves the fact.

In the complex case we already know that two $(n\times n)$ -matrices are conjugate if and only if they have the same Jordan normal form. The theorem tells us that this is also sufficient to determine if two real $(n\times n)$ -matrices are conjugate. The proof of the theorem shows us how, if this is the case, a real conjugating matrix $S$ can be found, starting with a complex conjugating matrix $T$ . Finding the Jordan form consists of finding a suitable basis for $V$ ; the complex conjugating matrix $T$ consists of the columns belonging to that basis.

In that way we can determine for each pair of $(n\times n)$ -matrices if they are conjugate, and even indicate a conjugating matrix. The real Jordan form of such a matrix, which we will treat later on, is (apart from a permutation of the Jordan blocks) a unique matrix from the conjugation class of $A$ .

Consider the $(2\times2)$ -matrix $A = \matrix{0 & -5 \\ 1 & 2 \\ }$ The minimal polynomial of $A$ is equal to the characteristic polynomial: $m_A(x) = p_A(x) =x^2-2 x+ 5$ The roots of this polynomial are $1-2 \complexi$ and $2 \complexi+1$ . The corresponding eigenvectors are $\matrix{5 \\ 2 \complexi-1 \\ }$ and $\matrix{5 \\ -2 \complexi-1 \\ }$ , respectively. The matrix $A$ is conjugate to the complex diagonal matrix $D = \matrix{1-2 \complexi &0\\ 0& 2 \complexi+1}$ The corresponding coordinate transformation is $T = \matrix{5 & 5 \\ 2 \complexi-1 & -2 \complexi-1 \\ }$ so $T^{-1}\, A\,T=D$ .

The matrices $T$ and $D$ are not real. If we regard $L_A$ , $L_D$ and $L_T$ as linear transformations of $\mathbb{C}^2$ , then the eigenvectors (the columns of $T$ ) each span a complex invariant subspace of $\mathbb{C}^2$ of dimension $1$ . By taking the real and imaginary part of the first eigenvector $\matrix{5 \\ 2 \complexi-1 \\ }$ , we find a basis for the original real vector space $\mathbb{R}^2$ relative to which $L_A$ has a real matrix. Determine that matrix.

$\matrix{1 & -2 \\ 2 & 1 \\ }$

By restricting scalars, we can view the linear map $L_A:\mathbb{C}^2\to \mathbb{C}^2$ as a linear map $\left.\left(L_D\right)\right|_{\mathbb{R}}:U\to U$ , where $U =\left.\mathbb{C}^2\right|_{\mathbb{R}}$ .

As a basis for $V$ we choose the real and imaginary part of the eigenvector $\matrix{5 \\ 2 \complexi-1 \\ }$ with eigenvalue $1-2 \complexi$ :
$\Re \matrix{5 \\ 2 \complexi-1 \\ } = \matrix{5 \\ -1 \\ } \phantom{xxx}\text{ and }\phantom{xxx} \Im \matrix{5 \\ 2 \complexi-1 \\ } =\matrix{0 \\ 2 \\ }$ The corresponding transition matrix is $S=\matrix{5 & 0 \\ -1 & 2 \\ }$ . The matrix of $A$ relative to said basis is
$\begin{array}{rcl} {S}^{-1}\,A\,S &=& {\matrix{{{1}\over{5}} & 0 \\ {{1}\over{10}} & {{1}\over{2}} \\ }}\, \matrix{0 & -5 \\ 1 & 2 \\ }\, \matrix{5 & 0 \\ -1 & 2 \\ }\\ &=& \matrix{1 & -2 \\ 2 & 1 \\ }\end{array}$

If we work with the eigenvector of $2 \complexi+1$ rather than the eigenvector of $1-2 \complexi$ , we get the transition matrix $\overline{S} = \matrix{5 & 0 \\ -1 & -2 \\ }$ . The matrix of $A$ with respect to the corresponding basis is
$\begin{array}{rcl} {\overline{S}}^{-1}\,A\,\overline{S} &=& {\matrix{{{1}\over{5}} & 0 \\ -{{1}\over{10}} & -{{1}\over{2}} \\ }}\, \matrix{0 & -5 \\ 1 & 2 \\ }\, \matrix{5 & 0 \\ -1 & -2 \\ }\\ &=& \matrix{1 & 2 \\ -2 & 1 \\ }\end{array}$ As the notation of the overline above $S$ suggests, this basis is the complex conjugate of the columns of $S$ .

The matrices ${S}^{-1}\,A\,S$ and ${\overline{S}}^{-1}\,A\,\overline{S}$ are conjugate by means of $\matrix{0&1\\ 1&0}$ :
$\matrix{1 & 2 \\ -2 & 1 \\ } =\matrix{0&1\\ 1&0} \matrix{1 & -2 \\ 2 & 1 \\ } \matrix{0&1\\ 1&0} = \matrix{0&1\\ 1&0}^{-1} \matrix{1 & -2 \\ 2 & 1 \\ } \matrix{0&1\\ 1&0}$ The matrices $S^{-1} A S$ and ${\overline{S}}^{-1}\,A\,\overline{S}$ are the real $(2\times2)$ -matrices with respect to the basis $\basis{1,\ii}$ that belong to complex multiplication by $1-2 \complexi$ and $2 \complexi+1$ , respectively.

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