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Free Physics Study Material for CSIR NET, IIT JAM, GATE, JEST & TIFR

by Pravegaa Education

Chapter 4: Matrices

Mathematical Physics, Part 1. Theory, numbered worked examples and 51 previous years’ questions (CSIR NET-JRF, GATE, IIT-JAM and JEST, 2020–2026) with full solutions. Click Show solution under any question.

Theory

Matrices

Outline

  1. Types of matrices: names and properties.

  2. Different operations on matrices (addition and multiplication).

  3. Eigenvalues and eigenvectors.

  4. Functions of matrices; diagonalization and similarity transformations.

Definition Matrices are rectangular arrays of numbers or functions. For example, \[ A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\qquad\text{or}\qquad A=\begin{bmatrix}\ee^x&\sin x\\x&\ee^{-x^2}\end{bmatrix}. \]

Applications

  • They store and process large amounts of data.

  • They express scientific and mathematical relations compactly.

  • They are the natural language of quantum mechanics.

Representation \[ \begin{bmatrix}\ &\ \\ \ &\ \end{bmatrix}_{m\times n}\Rightarrow m=\text{rows},\ n=\text{columns}; \qquad \begin{bmatrix}a_{11}&b_{12}\\c_{21}&d_{22}\end{bmatrix}_{2\times2}\Rightarrow\text{rows}=2,\ \text{columns}=2. \]

Operations on Matrices

(a) Addition and subtraction (only between matrices of equal dimensions): \[ \begin{bmatrix}a&b\\c&d\end{bmatrix}_{2\times2}\pm\begin{bmatrix}1&2\\3&4\end{bmatrix} =\begin{bmatrix}(a\pm1)&(b\pm2)\\(c\pm3)&(d\pm4)\end{bmatrix}. \]

(b) Multiplication \[ \underbrace{\begin{bmatrix}a&b\\c&d\end{bmatrix}}_{2\times2\ (m\times n)} \underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{2\times1\ (n\times p)} =\underbrace{\begin{bmatrix}ax+by\\cx+dy\end{bmatrix}}_{2\times1\ (m\times p)} =x\begin{bmatrix}a\\c\end{bmatrix}+y\begin{bmatrix}b\\d\end{bmatrix}. \]

This is a linear transformation: the matrix \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\) acts on the vector \(\begin{bmatrix}x\\y\end{bmatrix}\), and the origin stays fixed.

Example 4.1

Find the product of the matrices \(A\) and \(B\).

Solution.

(a) \(A=\begin{bmatrix}3&5&-1\\4&0&2\\-6&-3&2\end{bmatrix}_{3\times3}\) and \(B=\begin{bmatrix}2&-2&3&1\\5&0&7&8\\9&-4&1&1\end{bmatrix}_{3\times4}\): \begin{align*} AB&=\begin{bmatrix}6+25-9&-6+4&9+35-1&3+40-1\\8+18&-8-8&12+2&4+2\\-12-15+18&12-8&-18-21+2&-6-24+2\end{bmatrix} =\begin{bmatrix}22&-2&43&42\\26&-16&14&6\\-9&4&-37&-28\end{bmatrix}. \end{align*} (b) \(A=\begin{bmatrix}1&1\\100&100\end{bmatrix}\) and \(B=\begin{bmatrix}-1&1\\1&-1\end{bmatrix}\):   \(AB=\begin{bmatrix}1&1\\100&100\end{bmatrix}\begin{bmatrix}-1&1\\1&-1\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\).

(c) \(A=\begin{bmatrix}3&6&1\end{bmatrix}_{1\times3}\) and \(B=\begin{bmatrix}1\\2\\4\end{bmatrix}_{3\times1}\):   \(AB=\begin{bmatrix}3&6&1\end{bmatrix}\begin{bmatrix}1\\2\\4\end{bmatrix}=\begin{bmatrix}19\end{bmatrix}_{1\times1}\).

(d) \(A=\begin{bmatrix}1\\2\\4\end{bmatrix}_{3\times1}\) and \(B=\begin{bmatrix}3&6&1\end{bmatrix}_{1\times3}\):   \(AB=\begin{bmatrix}1\\2\\4\end{bmatrix}\begin{bmatrix}3&6&1\end{bmatrix}=\begin{bmatrix}3&6&1\\6&12&2\\12&24&4\end{bmatrix}_{3\times3}\).

(e) \(A=\begin{bmatrix}-1&1\\-1&-1\end{bmatrix}\) and \(B=\begin{bmatrix}1&1\\100&100\end{bmatrix}\): \[ AB=\begin{bmatrix}-1&1\\-1&-1\end{bmatrix}\begin{bmatrix}1&1\\100&100\end{bmatrix}=\begin{bmatrix}99&99\\-101&-101\end{bmatrix}, \] \[ BA=\begin{bmatrix}1&1\\100&100\end{bmatrix}\begin{bmatrix}-1&1\\-1&-1\end{bmatrix}=\begin{bmatrix}-2&0\\-200&0\end{bmatrix}, \] so \(AB\neq BA\).

Names of Matrices and their Meaning

1. Square matrices

Number of rows \(=\) number of columns. For example, \[ P=\begin{bmatrix}E_1&E_2&E_3\\E_1'&E_2'&E_3'\\E_1''&E_2''&E_3''\end{bmatrix}_{3\times3},\qquad m=n. \]

Transpose \(A^T\) Let \[ A=\begin{bmatrix}E_1&E_2&E_3\\E_4&E_5&E_6\\E_7&E_8&E_9\end{bmatrix}\;\Rightarrow\; A^T=\begin{bmatrix}E_1&E_4&E_7\\E_2&E_5&E_8\\E_3&E_6&E_9\end{bmatrix}. \] We write \(A=[a_{ij}]_{m\times n}\); for example \(a_{11}=E_1\), \(a_{12}=E_2\), \(a_{31}=E_7\), etc.

(a) \((A^T)^T=A\);    (b) \((AB)^T=B^TA^T\);    (c) \((A+B)^T=A^T+B^T\).

2. Symmetric matrix: \(A^T=A\)

\[ A=\begin{bmatrix}20&120&200\\120&10&150\\200&150&30\end{bmatrix},\qquad A^T=\begin{bmatrix}20&120&200\\120&10&150\\200&150&30\end{bmatrix}=A. \] \(\therefore A\) is a symmetric matrix.

3. Skew-symmetric matrix: \(A^T=-A\)

\[ B=\begin{bmatrix}0&1&-3\\-1&0&-2\\3&2&0\end{bmatrix},\qquad B^T=\begin{bmatrix}0&-1&3\\1&0&2\\-3&-2&0\end{bmatrix}=-\begin{bmatrix}0&1&-3\\-1&0&-2\\3&2&0\end{bmatrix}. \] Since \(B^T=-B\), \(B\) is skew-symmetric. The diagonal elements of a skew-symmetric matrix are always zero.

Proof If \(A^T=-A\), then \(a_{ji}+a_{ij}=0\). For \(i=j\), \(2a_{ii}=0\Rightarrow a_{ii}=0\).

4. Orthogonal matrix: \(A^TA=I A^T=A^-1</h3> \)

The sum of the squares of the elements of any row equals \(1\): \[ \text{if }\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\text{ is orthogonal}\;\Rightarrow\; a^2+b^2+c^2=1,\quad d^2+e^2+f^2=1,\quad g^2+h^2+i^2=1. \]

Symmetric and skew-symmetric parts Any square matrix can be written as \[ A=\frac12(A+A^T)+\frac12(A-A^T). \] With \(B=\frac12(A+A^T)\) and \(C=\frac12(A-A^T)\), \(A=B+C\), where \(B^T=B\) is symmetric and \(C^T=\frac12(A^T-A)=-C\) is skew-symmetric.

Example 4.2

If \(P=\begin{bmatrix}a\\b\end{bmatrix}\begin{bmatrix}a&-2b\end{bmatrix}\), what is its symmetric part?

Solution.

\(P=\begin{bmatrix}a\\b\end{bmatrix}\begin{bmatrix}a&-2b\end{bmatrix}=\begin{bmatrix}a^2&-2ab\\ba&-2b^2\end{bmatrix}\). The symmetric part is \[ B=\frac12(P+P^T)=\frac12\left\{\begin{bmatrix}a^2&-2ab\\ba&-2b^2\end{bmatrix}+\begin{bmatrix}a^2&ab\\-2ab&-2b^2\end{bmatrix}\right\} =\frac12\begin{bmatrix}2a^2&-ab\\-ab&-4b^2\end{bmatrix} =\begin{bmatrix}a^2&-ab/2\\-ab/2&-2b^2\end{bmatrix}. \]

5. Hermitian matrix: \(A^\dagger=A\)

For example, \(A=\begin{bmatrix}4&1-3\ii\\1+3\ii&7\end{bmatrix}\) has \(A^T=\begin{bmatrix}4&1+3\ii\\1-3\ii&7\end{bmatrix}\); complex conjugation gives \(A^\dagger=\begin{bmatrix}4&1-3\ii\\1+3\ii&7\end{bmatrix}=A\), so \(A\) is Hermitian.

6. Skew-Hermitian matrix: \(A^\dagger=-A\)

For example, \(A=\begin{bmatrix}3\ii&2+\ii\\-2+\ii&-\ii\end{bmatrix}\): \[ A^\dagger=\begin{bmatrix}-3\ii&-2-\ii\\2-\ii&\ii\end{bmatrix}=-\begin{bmatrix}3\ii&2+\ii\\-2+\ii&-\ii\end{bmatrix},\qquad A^\dagger=-A. \] \(\therefore A\) is a skew-Hermitian matrix.

7. Unitary matrix: \(A^\dagger A=I\) or \(A^=A^-1</h3> \)

8. Triangular matrix

If all elements below the principal diagonal are zero, as in \(\begin{bmatrix}a&b\\0&c\end{bmatrix}\), the matrix is upper triangular; if all elements above it are zero, it is lower triangular.

9. Diagonal matrix

For example, \(A=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}\): all off-diagonal elements are zero, and at least one diagonal element is non-zero.

10. Unit matrix

For example, \(A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\): \(1\)s on the principal diagonal, zeros elsewhere. It is also called the identity matrix.

11. Rotation matrices

Rotation matrix in 2D \[ A=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix},\] \[\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}x_1\\y_1\end{bmatrix}=\begin{bmatrix}x_1'\\y_1'\end{bmatrix}. \]

Rotation matrix in 3D \(+\theta\) for the anticlockwise direction; \(-\theta\) for the clockwise direction. \begin{gather*} A_x(\pm\theta)=\begin{bmatrix}1&0&0\\0&\cos\theta&\mp\sin\theta\\0&\pm\sin\theta&\cos\theta\end{bmatrix},\\ A_y(\pm\theta)=\begin{bmatrix}\cos\theta&0&\pm\sin\theta\\0&1&0\\ \mp\sin\theta&0&\cos\theta\end{bmatrix},\\ A_z(\pm\theta)=\begin{bmatrix}\cos\theta&\mp\sin\theta&0\\ \pm\sin\theta&\cos\theta&0\\0&0&1\end{bmatrix}. \end{gather*}

Elementary Row Transformations

Uses (a) Inverse; (b) linear equations; (c) rank.

Allowed operations

  1. interchange of rows;

  2. addition of a constant multiple of one row to another;

  3. multiplication of a row by a non-zero constant.

For example, for the system \begin{align*} 3x_1+2x_2+2x_3-5x_4&=8,\\ 0.6x_1+1.5x_2+1.5x_3-5.4x_4&=2.7,\\ 1.2x_1-0.2x_2-0.3x_3+2.4x_4&=2.1. \end{align*} the augmented matrix is \[ A=\begin{bmatrix}3&2&2&-5&8\\0.6&1.5&1.5&-5.4&2.7\\1.2&-0.2&-0.3&2.4&2.1\end{bmatrix}. \] Then \[ A\sim\begin{bmatrix}3&2&2&-5&8\\0&1.1&1.1&-4.4&1.1\\0&-1&-1.1&4.4&-1.1\end{bmatrix} \quad\begin{pmatrix}R_2\to R_2-0.2R_1\\R_3\to R_3-0.4R_1\end{pmatrix}, \] \[ A\sim\begin{bmatrix}3&2&2&-5&8\\0&1.1&1.1&-4.4&1.1\\0&0.1&0&0&0\end{bmatrix} \quad(R_3\to R_3+R_2). \]

Linear Dependence and Independence

Consider row vectors such as \(\vec A=[A_1\ A_2\ A_3]\), \(\vec B=[B_1\ B_2\ B_3]\), and a set \[ a_{(1)},a_{(2)},a_{(3)},\dots,a_{(m)}.\tag{A} \] A linear combination of them is \(c_1a_{(1)}+c_2a_{(2)}+\cdots+c_ma_{(m)}\). Let \[ c_1a_{(1)}+c_2a_{(2)}+c_3a_{(3)}+\cdots+c_ma_{(m)}=0.\tag{1} \]

  1. If (1) holds only when all \(c_i=0\), the vectors (A) are linearly independent.

  2. If, say, \(c_1a_{(1)}+c_2a_{(2)}=0\) with \(c_2\neq0\), then \(a_{(2)}=-c_1a_{(1)}/c_2\): the vectors are linearly dependent, since one is a combination of the others.

None of a set of linearly independent vectors can be dropped without losing information.

Example 4.3

Are \(a_{(1)}=\{3\ \ 0\ \ 2\ \ 2\}\), \(a_{(2)}=\{-6\ \ 42\ \ 24\ \ 54\}\), \(a_{(3)}=\{21\ \ -21\ \ 0\ \ -15\}\) linearly independent?

Solution.

\(6a_{(1)}=\{18\ \ 0\ \ 12\ \ 12\}\), \(-\frac12a_{(2)}=\{3\ \ {-21}\ \ {-12}\ \ {-27}\}\), \(-a_{(3)}=\{-21\ \ 21\ \ 0\ \ 15\}\), and these add to zero: \(6a_{(1)}-\frac12a_{(2)}-a_{(3)}=0\) with non-zero coefficients. So the three vectors are linearly dependent, while \(a_{(1)}\), \(a_{(2)}\) alone are independent (they are not proportional).

The Rank of a Matrix

The rank of a matrix \(A\) is the maximum number of linearly independent row vectors of \(A\). Let \[ A=\begin{bmatrix}3&0&2&2\\-6&42&24&54\\21&-21&0&-15\end{bmatrix}. \] Its rank is \(2\): the first two rows are independent and the third is a combination of them. By echelon form: \[ A\sim\begin{bmatrix}3&0&2&2\\0&42&28&58\\0&-21&-14&-29\end{bmatrix}\quad\begin{pmatrix}R_2\to R_2+2R_1\\R_3\to R_3-7R_1\end{pmatrix}, \] \[ A\sim\begin{bmatrix}3&0&2&2\\0&42&28&58\\0&0&0&0\end{bmatrix}\quad\Bigl(R_3\to R_3+\tfrac12R_2\Bigr). \] \(\therefore\) rank \(=2\).

  • \(A\) and \(A^T\) have the same rank.

  • For an \(n\times n\) matrix, rank \(=n\) if its \(n\) rows are linearly independent and rank \(<n\) if they are dependent.

The Inverse of a Matrix

  1. For a square \((n\times n)\) matrix \(A\), the inverse satisfies \(A^{-1}A=AA^{-1}=I\).

  2. \(A^{-1}=\dfrac{\operatorname{adj}A}{\abs A}\), which requires \(\abs A\neq0\); here \(\operatorname{adj}(A)=(\text{cofactor } A)^T\).

Eigenvalues and Eigenvectors

For a square matrix, \(A_{n\times n}X_{n\times1}=\lambda X_{n\times1}\), where \(\lambda\) is an eigenvalue. Then \[ (A-\lambda I)X=0\;\Rightarrow\;\abs{A-\lambda I}=0\quad\text{(for a non-trivial $X$)}. \] \(f(\lambda)\equiv\abs{A-\lambda I}=0\) is the characteristic equation. For a \(3\times3\) matrix \(f(\lambda)\) is a cubic with roots (eigenvalues) \(\lambda_1,\lambda_2,\lambda_3\) and corresponding eigenvectors \(X_1,X_2,X_3\).

Note

Every matrix satisfies its own characteristic equation (Cayley–Hamilton theorem).

Example 4.4

If \(A=\begin{bmatrix}-5&2\\2&-2\end{bmatrix}\), find the eigenvalues, the inverse and the eigenvectors of the matrix \(A\).

Solution.

Eigenvalues: \[ \abs{A-\lambda I}=0\;\Rightarrow\;\begin{vmatrix}-5-\lambda&2\\2&-2-\lambda\end{vmatrix}=0\;\Rightarrow\; (-5-\lambda)(-2-\lambda)-4=0\;\Rightarrow\;\lambda^2+7\lambda+6=0, \] so \(\lambda_1=-6\) and \(\lambda_2=-1\) are the eigenvalues.

Inverse. By the Cayley–Hamilton theorem, \(A^2+7A+6I=0\Rightarrow A^2=-[7A+6I]\). Multiplying by \(A^{-1}\), \[ A^{-1}(A^2+7A+6I)=0\;\Rightarrow\;A+7I+6A^{-1}=0\;\Rightarrow\;A^{-1}=-\frac16[A+7I]=-\frac16\begin{bmatrix}2&2\\2&5\end{bmatrix}. \] Or, directly: for \(A=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\), \(A^{-1}=\dfrac1{\det A}\begin{bmatrix}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{bmatrix}\); here \(\det A=10-4=6\), so \(A^{-1}=\dfrac16\begin{bmatrix}-2&-2\\-2&-5\end{bmatrix}\).

Eigenvectors. For \(\lambda_1=-6\): \[ [A+6I]X=\begin{bmatrix}1&2\\2&4\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=0\;\Rightarrow\;x_1+2x_2=0,\quad2x_1+4x_2=0\;\Rightarrow\;x_1=-2x_2. \] Let \(x_2=1\); then \(x_1=-2\). So \(X_1=\begin{bmatrix}-2\\1\end{bmatrix}\), with normalized form \(X_1=\dfrac1{\sqrt5}\begin{bmatrix}-2\\1\end{bmatrix}\).

For \(\lambda_2=-1\): \[ [A+I]Y=\begin{bmatrix}-4&2\\2&-1\end{bmatrix}\begin{bmatrix}y_1\\y_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\;\Rightarrow\; -4y_1+2y_2=0,\ 2y_1-y_2=0\;\Rightarrow\;y_2=2y_1. \] Let \(y_1=1\); then \(y_2=2\). So \(X_2=\begin{bmatrix}1\\2\end{bmatrix}\), with normalized form \(X_2=\dfrac1{\sqrt5}\begin{bmatrix}1\\2\end{bmatrix}\).

Since \(A^T=A\) is symmetric, its eigenvectors are orthogonal: \[ X_1^TX_2=\frac1{\sqrt5}\begin{bmatrix}-2&1\end{bmatrix}\frac1{\sqrt5}\begin{bmatrix}1\\2\end{bmatrix}=\frac15[-2+2]=0. \]

Location of eigenvalues

figureLocation of eigenvalues in the Argand plane: real for symmetric/Hermitian matrices, purely imaginary for skew-symmetric/skew-Hermitian matrices, and on the unit circle for orthogonal/unitary matrices.

Example 4.5

For \(A=\begin{bmatrix}-2&2&-3\\2&1&-6\\-1&-2&0\end{bmatrix}\), find the eigenvalues, the determinant, the trace, and the eigenvectors corresponding to the eigenvalues.

Solution.

\[ \abs{A-\lambda I}=0\;\Rightarrow\;\begin{vmatrix}-2-\lambda&2&-3\\2&1-\lambda&-6\\-1&-2&0-\lambda\end{vmatrix}=0 \;\Rightarrow\;\lambda^3+\lambda^2-21\lambda-45=0, \] i.e. \((\lambda+3)^2(\lambda-5)=0\), so the eigenvalues are \(\lambda_1=5\), \(\lambda_2=\lambda_3=-3\). By the Cayley–Hamilton theorem, \[ A^3+A^2-21A-45I=0\;\Rightarrow\;A^3+A^2-21A=45I=\begin{bmatrix}45&0&0\\0&45&0\\0&0&45\end{bmatrix}. \] With eigenvalues \(5,-3,-3\): \[ \Tr(A)=\sum\lambda_i=5-3-3=-1,\qquad \det(A)=\prod\lambda_i=(5)(-3)(-3)=45. \] Eigenvector for \(\lambda=5\): \[ [A-5I]X_1=\begin{bmatrix}-7&2&-3\\2&-4&-6\\-1&-2&-5\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}. \] Row reduction gives \(x_2+2x_3=0\) and \(x_1-2x_2-3x_3=0\). With \(x_3=k\): \(x_2=-2k\), \(x_1=2x_2+3x_3=-k\), so \(X_1=k(-1,-2,1)^T\). Taking \(k=-1\), \(X_1=\begin{bmatrix}1\\2\\-1\end{bmatrix}\); normalized, \(X_1=\dfrac1{\sqrt6}\begin{bmatrix}1\\2\\-1\end{bmatrix}\).

Eigenvectors for the degenerate eigenvalue \(\lambda_2=-3\): \[ [A+3I]X=\begin{bmatrix}1&2&-3\\2&4&-6\\-1&-2&3\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=0 \;\sim\;\begin{bmatrix}1&2&-3\\0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=0, \] so \(x_1+2x_2-3x_3=0\). With \(x_1=k_1\), \(x_2=k_2\): \(x_3=\frac13(k_1+2k_2)\), i.e. \(X=\bigl(k_1,\ k_2,\ \tfrac13(k_1+2k_2)\bigr)^T\). For \(k_1=k_2=1\), \(X_2=(1,1,1)^T\), normalized \(X_2=\frac1{\sqrt3}(1,1,1)^T\). A second, independent eigenvector (\(k_1=1\), \(k_2=0\)) is \[ X_3=\begin{bmatrix}1\\0\\1/3\end{bmatrix}\propto\begin{bmatrix}3\\0\\1\end{bmatrix}. \]

Pauli Spin Matrices

\[ \sigma_1=\begin{bmatrix}0&1\\1&0\end{bmatrix},\qquad \sigma_2=\begin{bmatrix}0&-\ii\\ \ii&0\end{bmatrix},\qquad \sigma_3=\begin{bmatrix}1&0\\0&-1\end{bmatrix}. \]

  1. \((\sigma_i)^2=I\): \[ (\sigma_1)^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I,\] \[(\sigma_2)^2=\begin{bmatrix}0&-\ii\\ \ii&0\end{bmatrix}\begin{bmatrix}0&-\ii\\ \ii&0\end{bmatrix}=I,\quad (\sigma_3)^2=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}=I. \]

  2. \(\abs{\sigma_i}=-1\) (the Pauli matrices are unitary and Hermitian).

  3. \(\sigma_i\sigma_j+\sigma_j\sigma_i=2\delta_{ij}I\), where \(\delta_{ij}\) is the Kronecker delta: \(\delta_{ij}=1\) if \(i=j\), \(\delta_{ij}=0\) if \(i\neq j\). For example, for \(i\neq j\), \[ \sigma_1\sigma_2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&-\ii\\ \ii&0\end{bmatrix}=\begin{bmatrix}\ii&0\\0&-\ii\end{bmatrix},\qquad \sigma_2\sigma_1=\begin{bmatrix}0&-\ii\\ \ii&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}-\ii&0\\0&\ii\end{bmatrix}, \] so \(\sigma_1\sigma_2+\sigma_2\sigma_1=0\); for \(i=j\), \(\sigma_1^2+\sigma_1^2=2I\).

  4. \(\sigma_j\sigma_k=\ii\sigma_l\) for \((j,k,l)\to(1,2,3),(2,3,1),(3,1,2)\); e.g. \(\sigma_1\sigma_2=\ii\sigma_3\).

Example 4.6

If \(P\) and \(Q\) are real symmetric, which of the following options is correct for \(PQ\)?

  1. symmetric for all \(P\) and \(Q\)
  2. never symmetric
  3. symmetric if \(PQ=QP\)
  4. antisymmetric for all \(P\) and \(Q\)
Solution.

(c). \((PQ)^T=Q^TP^T=QP\), which equals \(PQ\) only if \(PQ=QP\).

Example 4.7

If \(M=\dfrac1{\sqrt2}\begin{bmatrix}\ii&1\\1&\ii\end{bmatrix}\), the eigenvalues of \(M\) are

  1. real and positive
  2. purely imaginary with modulus 1
  3. complex with modulus 1

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