ExamplesB

Section 2.4 ExamplesB

Example 2.4.1.

If $A$ and $B$ are Hermitian, show that $AB+BA$ is Hermitian and $AB-BA$ is skew Hermitian.

Solution.

$\because$ $A$ and $B$ are Hermitian, we have

\begin{equation*} A^{\dagger}=A \quad \text{and} \quad B^{\dagger}=B. \end{equation*}

Hence,

\begin{equation*} \left(AB+BA\right)^{\dagger} = (AB)^{\dagger} + (BA)^{\dagger} \end{equation*}

\begin{equation*} = B^{\dagger}A^{\dagger}+A^{\dagger}B^{\dagger} =BA+AB=AB+BA, \end{equation*}

i.e., $AB+BA$ is Hermitian.

Again,

\begin{equation*} \left(AB-BA\right)^{\dagger} = (AB)^{\dagger} - (BA)^{\dagger} \end{equation*}

\begin{equation*} = B^{\dagger}A^{\dagger}-A^{\dagger}B^{\dagger} =BA-AB =-(AB-BA), \end{equation*}

i.e., $AB-BA$ is skew-Hermitian.

Example 2.4.2.

Show that the inverse of a matrix is unique.

Solution.

Let $A$ be a square matrix and the matrices $B$ and $C$ are inverse of $A\text{,}$ then $AB=BA=I$ also, $AC =CA=I\text{.}$ Now,

\begin{equation*} CAB=C(AB) = CI =C \quad \text{also}, \quad CAB = (CA)B=IB = B, \end{equation*}

That is,

\begin{equation*} CAB=B=C. \end{equation*}

Hence, $B$ is not different from $C\text{,}$ which implies that the inverse of a matrix is unique, i.e., there exists only one inverse matrix to a given matrix.

Example 2.4.3.

If $A$ is a non - singular matrix, then the transpose (conjugate transpose) of inverse is the inverse of the transpose (conjugate transpose) of $A\text{.}$

i.e., $(A^{-1})^{t} = (A^{t})^{-1}$ and $(A^{-1})^{\dagger}=(A^{\dagger})^{-1}\text{.}$

Solution.

we know that -

\begin{equation*} AA^{-1}=A^{-1}A=I \end{equation*}

\begin{equation*} \text{or,}\quad (AA^{-1})^{t}=(A^{-1}A)^{t}=(I)^{t}=I \end{equation*}

\begin{equation*} \text{or,}\quad (A^{-1})^{t}A^{t} = A^{t}(A^{-1})^{t} = I \end{equation*}

which follows that $(A^{-1})^{t}$ is the inverse of $A^{t}\text{.}$

i.e. $(A^{t})^{-1}=(A^{-1})^{t}\text{.}$ proved. Similarly other can be proved.

Example 2.4.4.

If $A$ and $B$ are any two square matrices of the same order, then show that

\begin{equation*} Tr(A+B)=TrA+TrB. \end{equation*}

Solution.

Let $A$ and $B$ are two matrices of the same order $n\text{.}$ Then, $Tr(A+B)$ = sum of the diagonal elements of the matrix $(A+B)$

\begin{equation*} = \sum\limits_{i=1}^{n}\left(a_{ij}+b_{ij}\right) =\sum\limits_{i=1}^{n}a_{ij}+\sum\limits_{i=1}^{n}b_{ij}=Tr(A)+Tr(B). \end{equation*}

Example 2.4.5.

Show that a real matrix is unitary if and only if it is orthogonal.

Solution.

If $A$is real matrix, then $A^{\dagger}=A^{t}$ for $A$ to be unitary, we have -

\begin{equation*} A^{\dagger}A=I \end{equation*}

\begin{equation*} \text{or,}\quad A^{t}A=I. \end{equation*}

$\therefore$ $A$ is orthogonal. [or, $A^{'}=A^{-1}\text{,}$ i.e., for a matrix to be orthogonal its transpose coincides with its inverse.]

Conversely, if $A$ is orthogonal, then

\begin{equation*} A^{t}A=I \end{equation*}

\begin{equation*} \text{or},\quad A^{\dagger}A =I \end{equation*}

$\therefore$ $A$ is unitary.

Example 2.4.6.

Show that under a unitary transformation, an orthonormal system of basis vectors is transformed into another orthonormal system.

Solution.

We require,

\begin{equation*} \delta_{ij}=\psi^{'}_{i} \psi^{'}_{j} = \sum\limits_{k} \gamma_{ki}\psi_{k}\cdot \sum\limits_{l} \gamma_{lj}\psi_{l} = \sum\limits_{kl}\gamma_{ki}^{*}\gamma_{lj}\psi_{k}\psi_{l} \end{equation*}

\begin{equation*} = \sum\limits_{kl}\gamma_{ki}^{*}\gamma_{lj}\delta_{kl} = \sum\limits_{kl}\gamma_{ki}^{*}\gamma_{kj} = \sum\limits_{k}(\gamma^{\dagger})_{ik} \gamma_{kj} \end{equation*}

here,

\begin{equation*} \delta_{ij}=\left(\gamma^{\dagger}\gamma\right)_{ij}. \end{equation*}

Thus, $\gamma^{\dagger}\gamma = 1 $ or, $\gamma$ must be an unitary matrix.

Example 2.4.7.

Show that a Hermitian matrix remains Hermitian under unitary transformation.

Solution.

For unitary transformation,

\begin{equation} AA^{\dagger}=A^{\dagger}A =I \tag{2.4.1} \end{equation}

If $A$ to be Hermitian, then $A^{\dagger}=A\text{.}$ Taking conjugate transpose of eqn. (2.4.1)>, we get -

\begin{equation*} \left(AA^{\dagger}\right)^{\dagger} = \left(A^{\dagger}A\right)^{\dagger} = I^{\dagger} \end{equation*}

\begin{equation*} \text{or,} \quad \left(A^{\dagger}\right)^{\dagger}A^{\dagger} = A^{\dagger}\left(A^{\dagger}\right)^{\dagger} =I \end{equation*}

[$\because I^{\dagger}=I$ is an identity matrix.]

\begin{equation*} \therefore AA^{\dagger}=A^{\dagger}A =I. \end{equation*}

This equality shows that Hermitian matrix remains Hermitian under unitary transformation.

Example 2.4.8.

Show that the length of a real vector is preserved under orthogonal transformation.

Solution.

Let $x_{i}$ and $y_{i}$ are the real vectors of matrices $X$ and $Y$ which is related as

\begin{equation} Y=AX \tag{2.4.2} \end{equation}

where $A$ is a transformation matrix. For an orthogonal transformation, we have -

\begin{equation} Y^{'}Y=(AX)^{'}(AX) = X^{'}A^{'}AX =X^{'}X \tag{2.4.3} \end{equation}

Since $A$ is considered to be an orthogonal. i.e., $A^{'}A =I\text{.}$

Therefore, from eqn. (2.4.3), we get -

\begin{equation*} y^{2}_{i} = x^{2}_{i} \end{equation*}

where,

\begin{equation} \sum\limits_{i=1}^{n}y_{i}=\sum\limits_{j=1}^{n}a_{ij}x_{j}; \tag{2.4.4} \end{equation}

\begin{equation*} Y' =\begin{bmatrix} \cdot & \cdot & \cdot \end{bmatrix}; \end{equation*}

\begin{equation*} \text{and}\quad Y=\begin{bmatrix} \cdot\\\cdot\\\cdot \end{bmatrix}. \end{equation*}

Note: In case $X$ and $X^{'}$ are complex then the orthogonal conditions to be

\begin{equation*} \left(AA^{\dagger}\right) = \left(A^{\dagger}A\right) = I, \end{equation*}

i.e., A is an unitary matrix. A real unitary matrix is an orthogonal matrix. (2.4.4) shows that the norm of a real vector remains invariant under orthogonal transformation. i.e.,

\begin{equation*} \Vert X \Vert = \Vert AX \Vert \end{equation*}

\begin{equation*} [\because \quad \Vert X \Vert = \sqrt{x^{2}_{i}} \quad \text{and}\quad \Vert Y \Vert = \sqrt{y^{2}_{i}}]. \end{equation*}

Example 2.4.9.

Show that the eigen values of a Hermitian operator is real and eigen vectors belonging to different eigen values are orthogonal. Or, The eigen values of a Hermitian matrix are all real.

Solution.

Let $\lambda_{i}$ and $\lambda_{j}$ be two eigen values and $X_{i}$ and $X_{j}$ be two eigen vectors of Hermitian matrix (operator) $H\text{.}$ Then

\begin{align} HX_{i} \amp = \lambda_{i}X_{i} \tag{2.4.5}\\ HX_{j} \amp = \lambda_{j}X_{j} \tag{2.4.6} \end{align}

premultiplying by $X^{\dagger}_{j}$ to eqn. (2.4.5) and $X^{\dagger}_{i}$ to eqn. (2.4.6), respectively, we get -

\begin{align} X^{\dagger}_{j} HX_{i} \amp = \lambda_{i}X^{\dagger}_{j}X_{i} \tag{2.4.7}\\ \text{and}\quad X^{\dagger}_{i} HX_{j} \amp = \lambda_{j}X^{\dagger}_{i}X_{j} \tag{2.4.8} \end{align}

taking conjugate transpose of eqn. (2.4.8), we get -

\begin{equation*} \left(X^{\dagger}_{i} HX_{j}\right)^{\dagger}=\left(\lambda_{j} X_{i}^{\dagger}X_{j}\right)^{\dagger} \end{equation*}

[for Hermitian matrix $H^{\dagger} =H$] or,

\begin{equation} X^{\dagger}_{j} HX_{i} = \bar{\lambda}_{j}X^{\dagger}_{j}X_{i} \tag{2.4.9} \end{equation}

equating eqns. (2.4.7) and (2.4.9), we get -

\begin{align} \lambda_{i}X^{\dagger}_{j}X_{i} \amp = \bar{\lambda}_{j}X^{\dagger}_{j}X_{i} \tag{2.4.10}\\ \left(\lambda_{i} - \bar{\lambda}_{j}\right)X^{\dagger}_{j}X_{i} \amp = 0\tag{2.4.11} \end{align}

Case I: Let $i=j\text{,}$ then $\left(\lambda_{i} - \bar{\lambda}_{i}\right)=0$ [$\because X^{\dagger}_{i}X_{i} \neq 0\text{,}$ as $X_{i}$ is a non - zero vector.

\begin{equation*} \therefore \quad \lambda_{i} = \bar{\lambda}_{i} \end{equation*}

i.e., $\lambda_{i}$ is real for each $i\text{.}$

Case II: Let $i\neq j\text{.}$ we have $\lambda_{i} \neq \bar{\lambda}_{j}\text{.}$

\begin{equation*} \therefore \quad X^{\dagger}_{j}X_{i} = 0. \end{equation*}

which means eigen vectors for different eigen values of $H$ are orthogonal.

Example 2.4.10.

Show that the trace and determinant of a matrix is invarinat under similarity transformation.

Solution.

Let $X$ and $Y$ are two vectors relative to the basis of unprimed system and are represented by $X'$ and $Y'$ relative to the basis of a primed system, so that

\begin{equation} X=PX'; \quad Y=PY' \tag{2.4.12} \end{equation}

where $P$ is transformation matrix. Suppose $X$ and $Y$ are themselves related by the transformation matrix $A$ in the unprimed system, so that

\begin{equation} X=AY\tag{2.4.13} \end{equation}

Now, from eqns. (2.4.12) and (2.4.13), we have

\begin{align} X' \amp = P^{-1}X = P^{-1}AY = P^{-1}APY' \equiv BY' \tag{2.4.14}\\ \text{where} \quad B \amp = P^{-1}AP \tag{2.4.15} \end{align}

This implies that

\begin{equation} A = PBP^{-1}.\tag{2.4.16} \end{equation}

i.e., the matrix of transformation for the same two vectors in the coordinate system is related by $P^{-1}AP$ and $PBP^{-1}\text{.}$ Hence $A$ and $B$ are said to be related under similarity transformation. From eqn. (2.4.15), we have

\begin{equation*} Tr(B)=\sum\limits_{i}B_{ii}=\sum\limits_{i}(P^{-1}AP)_{ii} =\sum\limits_{i}\sum\limits_{jk}P^{-1}_{ij}A_{jk}P_{ki} \end{equation*}

\begin{equation*} = \sum\limits_{j}\sum\limits_{k}\left[\sum\limits_{i}P_{ki}P^{-1}_{ii}\right]A_{jk} \end{equation*}

\begin{equation*} \left[\because \quad \sum\limits_{i}(AB)_{ij}=\sum\limits_{i}\sum\limits_{j}a_{ij}b_{ji}\right] \end{equation*}

after changing the sign for the places.

\begin{equation*} =\sum\limits_{j}\sum\limits_{k}\left[\sum\limits_{i}PP^{-1}\right]_{kj}A_{jk} = \sum\limits_{j}\left[APP^{-1}\right]_{jj}= \sum\limits_{j}A_{jj} = Tr(A) \end{equation*}

i.e., trace is invarient under similarity transformation. Also,

\begin{equation*} |B| = det B = [P^{-1}AP| = |P^{-1}||A||P| = A \end{equation*}

[$\because |P^{-1}||P| = 1$] proved.

Example 2.4.11.

Show that diagonalizing matrix of a Hermitian matrix is unitary.

Solution.

Let $H$ be a hermitian matrix and $R$ be its diagonalizing matrix, then

\begin{equation} R^{-1}HR=diag\left(\lambda_{1}, \lambda_{2}, \lambda_{3}, \cdots, \lambda_{n} \right)\tag{2.4.17} \end{equation}

where $\lambda_{1}{,} \lambda_{2}{,} \lambda_{3}{,} \cdots{,} \lambda_{n}$ are the eigen values of $H$ and are all real.

Taking transposed conjugate on both sides of eqn. (2.4.17), we have

\begin{align*} \left(R^{-1}HR\right)^{\dagger} \amp =\left[diag\left(\lambda_{1}, \lambda_{2}, \lambda_{3}, \cdots, \lambda_{n} \right)\right]^{\dagger} = diag\left(\lambda_{1}, \lambda_{2}, \lambda_{3}, \cdots, \lambda_{n} \right)\\ \text{or,}\quad R^{\dagger}H(R^{-1})^{\dagger} \amp = diag\left(\lambda_{1}, \lambda_{2}, \lambda_{3}, \cdots, \lambda_{n} \right) = R^{-1}HR \end{align*}

which gives

\begin{equation*} R^{\dagger} = R^{-1} \end{equation*}

\begin{equation*} \text{or,}\quad RR^{\dagger} = RR^{-1} = I \end{equation*}

$\therefore$ R is a unitary matrix.

Example 2.4.12.

Use matrix method to find the voltage drop at resistor $R_{6} \text{.}$ In the given figure $R_{1} = 1\Omega \text{,}$ $R_{2} = 2\Omega$ , $R_{3}=3\Omega\text{,}$ $R_{4} = 4\Omega\text{,}$ $R_{5}=5\Omega\text{,}$ $R_{6}=6\Omega\text{,}$ $E_{1} =5 Volt\text{,}$ and $E_{2} =10 Volt\text{.}$

Solution.

Consider $I_{1}{,} I_{2}{,} I_{3}$ are the currents flowing in the loops as shown in Figure 2.4.13 Now using Kirchhoff’s voltage law $\sum IR = \sum V\text{,}$ from loop 1, we have

\begin{align*} 1.I_{1}+2.I_{1}+3.I_{1}-3.I_{2} \amp = 5 \\ \text{or,}\quad 6.I_{1}-3.I_{2} + 0.I_{3} \amp = 5 \end{align*}

From loop 2,

\begin{align*} 5. I_{2}+3.I_{2}+4.I_{2}-3.I_{1}+5.I_{3} \amp = 10 \\ \text{or,}\quad -3.I_{1}+12.I_{2} +5.I_{3} \amp = 10 \end{align*}

From loop 3,

\begin{align*} 6.I_{3}+5.I_{3}+5.I_{2} \amp = 5 \\ \text{or,}\quad 0.I_{1}+5.I_{2} +11.I_{3} \amp = 5 \end{align*}

Hence in the matrix form

\begin{equation*} \begin{bmatrix} 6 & -3 & 0 \\ -3 & 12 & 5\\ 0 & 5 & 11 \end{bmatrix} \begin{bmatrix} I_{1}\\ I_{2}\\I_{3} \end{bmatrix} = \begin{bmatrix} 5\\ 10\\5 \end{bmatrix} \end{equation*}

The determinant of the given matrix is

\begin{equation*} D=\begin{vmatrix} 6 & -3 & 0 \\ -3 & 12 & 5\\ 0 & 5 & 11 \end{vmatrix} \end{equation*}

\begin{equation*} = 6(12\times 11-5\times 5)+3(-3\times 11-0\times 5) +0(-3\times 5-0\times 12) = 543. \end{equation*}

\begin{equation*} \therefore \quad I_{1} = \frac{\begin{vmatrix} 5 & -3 & 0 \\ 10 & 12 & 5\\ 5 & 5 & 11 \end{vmatrix}}{D} = 1.455 A \end{equation*}

\begin{equation*} I_{2} = \frac{\begin{vmatrix} 6 & 5 & 0 \\ -3 & 10 & 5\\ 0 & 5 & 11 \end{vmatrix}}{D} = 1.243 A \end{equation*}

\begin{equation*} I_{3} = \frac{\begin{vmatrix} 6 & -3 & 5 \\ -3 & 12 & 10\\ 0 & 5 & 5 \end{vmatrix}}{D} = -0.1105 A \end{equation*}

$-ve$ sign indicats that the current $I_{3}$ should be taken in opposite direction. Hence, the potential drop at $R_{6} = 6\times 0.1105 = 0.663$ volt.

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