Square Matrix as a Linear Operator

For a given matrix \(\,\boldsymbol{A}\,\in M_n(C)\,\) we define a mapping

(1)\[F:\quad C^n \ni \,\boldsymbol{x} \ \,\to\ \, F(\boldsymbol{x})\,:\,=\, \boldsymbol{A}\,\boldsymbol{x} \,\in\, C^n\,.\]

More precisely, \(\,\) if \(\ \ \boldsymbol{A}= [\,\boldsymbol{A}_1\,|\,\boldsymbol{A}_2\,|\,\ldots\,|\,\boldsymbol{A}_n\,]= [a_{ij}]_{n\times n}\,,\ \ \boldsymbol{x}\ =\ [x_{i}]_n\,,\ \) then

\[\begin{split}F\ \left[\begin{array}{c} x_{1} \\ x_{2} \\ \ldots \\ x_{n} \end{array}\right]\ = \ \left[\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \ldots & \ldots & \ldots & \ldots \\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{array}\right]\ \left[\begin{array}{c} x_{1} \\ x_{2} \\ \ldots \\ x_{n} \end{array}\right]\,.\end{split}\]

We take a simplified notation for an image of a vector \(\,\boldsymbol{x}\,\) under the mapping \(\,F\):

\[F(\boldsymbol{x})\rightarrow F\boldsymbol{x}\,.\]

Theorem 0. \(\\ \ \) The mapping (1) is a linear operator on the space \(\,C^n\,\): \(\ \ F\in\text{End}\,(C^n)\).

Proof. \(\\\) Linearity of the mapping \(F\) is a consequence of matrix multiplication:

\[ \begin{align}\begin{aligned}F(\boldsymbol{x}+\boldsymbol{y})\ =\ \boldsymbol{A}\,(\boldsymbol{x}+\boldsymbol{y})\ =\ \boldsymbol{A}\,\boldsymbol{x}\ +\ \boldsymbol{A}\,\boldsymbol{y}\ =\ F(\boldsymbol{x})\, +\, F(\boldsymbol{y})\,,\\F(\lambda\,\boldsymbol{x})\ =\ \boldsymbol{A}\,(\lambda\,\boldsymbol{x})\ =\ \lambda\ (\boldsymbol{A}\,\boldsymbol{x})\ =\ \lambda\ F(\boldsymbol{x})\,, \quad \boldsymbol{x},\boldsymbol{y}\,\in C^n,\ \ \lambda \in C.\end{aligned}\end{align} \]

Theorem 1. \(\\ \ \) \(\boldsymbol{A}\,\) is a matrix of the operator \(\,F\,\) in the canonical basis \(\,\mathcal{E}\,\) of the space \(\,C^n :\) \(\ \boldsymbol{A}\,=\,M_{\mathcal{E}}(F)\).

Proof. \(\\\) Let \(\,\mathcal{E}\,=\,(\boldsymbol{e}_1,\, \boldsymbol{e}_2,\,\dots,\,\boldsymbol{e}_n)\,\) be the canonical basis of the space \(\,C^n.\ \) Then

\[\begin{split}\begin{array}{rl} M_{\mathcal{E}}(F) \!\!\! & =\ \ [\ F(\boldsymbol{e}_1)\ |\ F(\boldsymbol{e}_2)\ | \ \ldots\,|\ F(\boldsymbol{e}_n)\,]\ \ = \\ & =\ \ [\ \boldsymbol{A}\boldsymbol{e}_1\,| \ \boldsymbol{A}\boldsymbol{e}_2\,| \,\ldots\,|\ \boldsymbol{A}\boldsymbol{e}_n\,]\ \ = \\ & = \ \ [\ \boldsymbol{A}_1\,|\ \boldsymbol{A}_2\,| \ \ldots\,|\ \boldsymbol{A}_n\,]\ \ =\ \ \boldsymbol{A}\,. \end{array}\end{split}\]

Theorem 2. \(\\ \ \) Eigenvalues of the operator \(\,F\,\) are roots of the characteristic equation of the matrix \(\boldsymbol{A}.\)

Proof. \(\\\) A vector \(\,\boldsymbol{x} \in C^n\setminus\{\boldsymbol{0}\}\,\) is an eigenvector of the operator \(\,F\,\) associated with the eigenvalue \(\,\lambda \in C\) \(\,\) if

(2)\[ \begin{align}\begin{aligned}\begin{split}\begin{array}{rc} & F(\boldsymbol{x})\,=\,\lambda\,\boldsymbol{x}, \\ \text{that is} & \boldsymbol{A}\,\boldsymbol{x} = \lambda\,\boldsymbol{x}\,, \end{array}\end{split}\\(\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)\ \boldsymbol{x}\ =\ \boldsymbol{0}\,.\end{aligned}\end{align} \]

The homogeneous linear problem (2) has a non-zero solution \(\,\boldsymbol{x}\,\) if and only if \(\\\) \(\ \det{(\boldsymbol{A}-\lambda\,\boldsymbol{I}_n)}\ =\ 0\,,\ \) that is, if \(\,\lambda\,\) is a characteristic root of the matrix \(\,\boldsymbol{A}\).

Numbers \(\,\lambda\,\) and the associated non-zero vectors \(\,\boldsymbol{x}\,\) determined by the equation (2) will be called eigenvalues and eigenvectors of the matrix \(\,\boldsymbol{A}\).

Hence, a vector \(\,\boldsymbol{x} \in C^n\,\) is an eigenvector of the matrix \(\,\boldsymbol{A} \in M_n(C)\ \) associated with an eigenvalue \(\,\lambda \in C\ \) if \(\,\boldsymbol{x}\neq\boldsymbol{0}\ \ \) and \(\ \ \boldsymbol{A}\,\boldsymbol{x} = \lambda\,\boldsymbol{x}\).

Theorem 3.

1. \(\ \) If \(\boldsymbol{A}\,\) is a Hermitian matrix, \(\ \) then \(\,F\,\) is a Hermitian operator:

   \[\boldsymbol{A}^+=\ \boldsymbol{A}\quad\Rightarrow\quad F^+=\ F\,.\]

2. \(\ \) If \(\boldsymbol{A}\,\) is a unitary matrix, \(\ \) then \(\,F\,\) is a unitary operator:

   \[\boldsymbol{A}^+\boldsymbol{A}\,=\,\boldsymbol{I}_n \quad\Rightarrow\quad F^+F\,=\,I\,,\]

   (\(I\ \) the identity operator on the space \(\ C^n\)).

3. \(\ \) If \(\boldsymbol{A}\,\) is a normal matrix, \(\ \) then \(\,F\,\) is a normal operator:

   \[\boldsymbol{A}^+\boldsymbol{A}\,=\,\boldsymbol{A}\,\boldsymbol{A}^+ \quad\Rightarrow\quad F^+F\,=\,F\,F^+\,.\]

Hence, all the theorems about eigenvalues and eigenvectors of Hermitian (unitary, normal) operators may be applied to eigenvalues and eigenvectors of Hermitian (unitary, normal) matrices.

Introduction to a proof of Theorem 3.

Hermitian conjugate of a matrix \(\boldsymbol{A} \in M_n(C)\):

\[\boldsymbol{A}^+:\,=\,(\boldsymbol{A}^T)^* =\,(\boldsymbol{A}^*)^T\,.\]

Hermitian conjugation of an operator \(\ F\in\text{End}(V)\ \) on a unitary space \(\ V=V(C)\ \):

\[\langle\boldsymbol{x},F^+\boldsymbol{y}\rangle:\,=\, \langle F\boldsymbol{x},\boldsymbol{y}\rangle\,,\quad \boldsymbol{x},\boldsymbol{y}\in V\,.\]

A necessary and sufficient condition for equality of two vectors:

Let \(\ \boldsymbol{x},\boldsymbol{y} \in V\,,\ \) where \(\ V=V(C)\ \) unitary. Then

\[\boldsymbol{x} = \boldsymbol{y} \quad \Leftrightarrow \quad \langle \boldsymbol{z}, \boldsymbol{x} \rangle = \langle \boldsymbol{z}, \boldsymbol{y} \rangle \quad \text{for all } \boldsymbol{z} \in V\,.\]

A necessary and sufficient condition for equality of two linear operators:

Let \(\ F,G\in\text{End}(V)\,,\ \) where \(\ V=V(C)\ \) unitary. Then

\[F = G \quad \Leftrightarrow \quad \langle \boldsymbol{x}, F \boldsymbol{y} \rangle = \langle \boldsymbol{x}, G \boldsymbol{y} \rangle \quad \text{for all } \boldsymbol{x},\boldsymbol{y} \in V\,.\]

An inner product on the space \(C^n\):

For \(\ \ \boldsymbol{x}\ =\ \left[\begin{array}{c} x_1 \\ x_2 \\ \ldots \\ x_n \end{array}\right],\ \ \boldsymbol{y}\ =\ \left[\begin{array}{c} y_1 \\ y_2 \\ \ldots \\ y_n \end{array}\right] \in C^n\,,\)

\[\begin{split}\langle \boldsymbol{x},\boldsymbol{y} \rangle \ =\ x_1^*\,y_1\,+\;x_2^*\,y_2\,+\;\dots\;+\;x_n^*\,y_n \,=\;[\,x_1^*,\,x_2^*,\,\dots,\,x_n^*\,]\ \left[\begin{array}{c} y_1 \\ y_2 \\ \dots \\ y_n \end{array}\right]\ =\ \boldsymbol{x}^+\boldsymbol{y}\,.\end{split}\]

Hermitian conjuagation \(\ F^+\ \) of the operator \(\ F\ \) given by (1) describes

Lemma.

(3)\[\begin{split}\begin{array}{lc} & F(\boldsymbol{x})=\boldsymbol{A}\,\boldsymbol{x}\quad\Rightarrow\quad F^+(\boldsymbol{x})=\boldsymbol{A}^+\boldsymbol{x}, \\ \text{that is,} & F\,\boldsymbol{x}=\boldsymbol{A}\,\boldsymbol{x}\quad\Rightarrow\quad F^+\,\boldsymbol{x}=\boldsymbol{A}^+\boldsymbol{x}. \end{array}\end{split}\]

Proof of the lemma. For every vector \(\ \boldsymbol{y}\in C^n:\)

\[\begin{split}\begin{array}{rcl} \langle\boldsymbol{y},F^+\boldsymbol{x}\rangle \!\! & =\ \ \ \langle F\boldsymbol{y},\boldsymbol{x}\rangle \ =\ \langle \boldsymbol{A}\,\boldsymbol{y},\boldsymbol{x}\rangle\ = & \\ & =\ (\boldsymbol{A}\,\boldsymbol{y})^+\,\boldsymbol{x}\ =\ \ \boldsymbol{y}^+\boldsymbol{A}^+\boldsymbol{x}\ = & \!\! \langle\boldsymbol{y},\boldsymbol{A}^+\boldsymbol{x}\rangle\,, \end{array}\end{split}\]

and thus \(\ F^+\boldsymbol{x}=\boldsymbol{A}^+\boldsymbol{x}, \ \ \boldsymbol{x}\in C^n\).

Proof of the Theorem 3.

1. \(\ \) Assume that \(\ \boldsymbol{A}^+=\,\boldsymbol{A}.\ \) Then for arbitrary \(\,\boldsymbol{x},\boldsymbol{y}\in C^n\):

   \[\begin{split}\begin{array}{rll} \langle\boldsymbol{x},F^+\boldsymbol{y}\rangle \!\! & =\ \ \langle F\boldsymbol{x},\boldsymbol{y}\rangle\ \ =\ \ \langle\boldsymbol{A}\,\boldsymbol{x},\boldsymbol{y}\rangle\ \ =\ \ (\boldsymbol{A}\,\boldsymbol{x})^+\boldsymbol{y}\ \ = & \\ & =\ \ \boldsymbol{x}^+\boldsymbol{A}^+\boldsymbol{y}\ \ =\ \ \ \boldsymbol{x}^+\boldsymbol{A}\,\boldsymbol{y}\ \ \ =\ \ \langle\boldsymbol{x},\boldsymbol{A}\,\boldsymbol{y}\rangle\ \ = & \!\! \langle\boldsymbol{x},F\boldsymbol{y}\rangle , \end{array}\end{split}\]

   so that \(\ F^+=\ F\).

2. \(\ \) Assume that \(\ \boldsymbol{A}^+\boldsymbol{A}=\boldsymbol{I}_n.\ \) Then for arbitrary \(\,\boldsymbol{x},\boldsymbol{y}\in C^n\):

   \[\begin{split}\begin{array}{rll} \langle\boldsymbol{x},(F^+F)\,\boldsymbol{y}\rangle \!\! & =\ \ \langle\boldsymbol{x},F^+(F\boldsymbol{y})\rangle\ \ =\ \ \langle F\boldsymbol{x}\,,F\boldsymbol{y}\rangle\ \ =\ \ \langle\boldsymbol{A}\,\boldsymbol{x}, \boldsymbol{A}\,\boldsymbol{y}\rangle\ \ = & \\ & =\ \ (\boldsymbol{A}\boldsymbol{x})^+\, (\boldsymbol{A}\boldsymbol{x})\ \ =\ \ \boldsymbol{x}^+\boldsymbol{A}^+ \boldsymbol{A}\,\boldsymbol{y}\ \ \, = \quad \boldsymbol{x}^+\boldsymbol{I}_n\,\boldsymbol{y}\quad\ = & \langle\boldsymbol{x},I\,\boldsymbol{y}\rangle , \end{array}\end{split}\]

   so that \(\ F^+F=I\).

3. \(\ \) Assume that \(\ \boldsymbol{A}^+\boldsymbol{A}= \boldsymbol{A}\boldsymbol{A}^+.\ \) By Lemma (3),

   \[\begin{split}\begin{array}{rl} \langle\boldsymbol{x},(F^+F)\,\boldsymbol{y}\rangle \!\! & =\ \ \langle\boldsymbol{x},F^+(F\boldsymbol{y})\rangle\ \ =\ \ \langle F\boldsymbol{x}\,,F\boldsymbol{y}\rangle\ \ =\ \ \langle\boldsymbol{A}\,\boldsymbol{x}, \boldsymbol{A}\,\boldsymbol{y}\rangle\ \ = \\ & =\ (\boldsymbol{A}\,\boldsymbol{x})^+ (\boldsymbol{A}\,\boldsymbol{y})\ \ =\ \ \boldsymbol{x}^+(\boldsymbol{A}^+\boldsymbol{A})\,\boldsymbol{y}\ \ =\ \ \boldsymbol{x}^+(\boldsymbol{A}\boldsymbol{A}^+)\,\boldsymbol{y}\ = \\ & =\ (\boldsymbol{A}^+\boldsymbol{x})^+ (\boldsymbol{A}^+\boldsymbol{y})\ =\ \langle\boldsymbol{A}^+\boldsymbol{x}, \boldsymbol{A}^+\boldsymbol{y}\rangle\ =\ \langle F^+\boldsymbol{x},F^+\boldsymbol{y}\rangle\ = \\ \langle\boldsymbol{x},(FF^+)\,\boldsymbol{y}\rangle , \end{array}\end{split}\]

   for every \(\,\boldsymbol{x},\boldsymbol{y}\in C^n,\ \) and thus \(\ F^+F=FF^+\).