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If $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero eigenvector, then $A^2\mathbf{v}={\lambda }^2\mathbf{v}=A\mathbf{v}=\lambda \mathbf{v}$. Thus $\lambda (\lambda -1)=0$, soBoth values need not occur: the zero matrix has only $0$, and the identity matrix only $1$. More precisely, an idempotent linear map whose rank of a linear map is $r$ has $r$ copies of $1$ and $n-r$ copies of $0$. Indeed, its image of a linear map and kernel of a linear map form a direct sum, giving an eigenbasis without assuming an orthogonal projection.

For $\mu \ne 0$, the adjugate matrix formula, or elimination on the three unit right-hand sides, givesMultiplication by $M$ gives the identity matrix, checking all signs and coefficients.

The matrix of cofactors is $\left(\begin{array}{ccc}-1& 2& -3\\ 2& 0& -2\\ -3& -2& 3\end{array}\right)$. Transposing and dividing by the nonzero determinant gives the inverse matrixMultiplication by $S$ verifies the identity matrix.

The permutation matrix acts by $A(x,y,z{)}^T=(y,z,x{)}^T$. Applying this cyclic permutation three times returns every vector to itself, proving $A^3=I$. Explicitly $A^2=\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$ and multiplication once more gives the identity matrix.

The adjugate matrix formula givesMultiplication by the right-side column yieldsMultiplying the inverse by $A$ checks the identity matrix, and substitution of this solution into the three original rows gives $1,2,3$ respectively.

Interchange the diagonal entries, negate the off-diagonal entries, and divide by the nonzero determinant. The matrix inverse isMultiplication by $A$ gives the identity matrix, since each diagonal product sum is $25-9=16$ and each off-diagonal product sum is zero.

The determinant is $2\cdot 4-3\cdot 1=5\ne 0$, so the matrix inverse exists. Using the two-by-two adjugate matrix formula givesMultiplication on either side gives the identity matrix, which also checks the signs of the off-diagonal entries.

Put $d=c-1$. The matrix is real and symmetric, so orthogonality means $A^{\mathsf{T}}A=A^2=I$. Multiplication gives the scalar conditionsIf $d=0$, then $c=1$ and $a=\pm 1$; if $a=0$, then $d=\pm 1$ and $c=0$ or $2$. In both cases $b=\pm 1$. The eight parameter triples are thereforeThese conditions are sufficient as well as necessary, since they make every entry of $A^{\mathsf{T}}A$ equal to the corresponding entry of the identity matrix.

In the nonsingular case the unique solution isGeometrically this is a single selected point of the straight line through $\mathbf{a}$ parallel to $\mathbf{c}$.

When $\mathbf{b}\cdot \mathbf{c}=1$ and $\mathbf{b}\cdot \mathbf{a}=0$, the solution set is that entire straight line, $\mathbf{a}+\lambda \mathbf{c}$. The remaining case has empty solution set. In matrix terms the equation is $(I-\mathbf{c}{\mathbf{b}}^T)\mathbf{x}=\mathbf{a}$, a rank-one matrix update of the identity matrix.

The similarity transformation gives $A=PD{P}^{-1}$. In the matrix power, each adjacent pair $P^{-1}P$ cancels to the identity matrix:This also follows by mathematical induction: the formula holds for $n=1$, and multiplication by $A=PD{P}^{-1}$ turns its $n$th instance into the $(n+1)$st.

A complex $3\times 3$ matrix is diagonalizable exactly when it has a basis of three linearly independent eigenvectors. If all its eigenvalues are one, a diagonalization would have $D=I$ and hence $A=PI{P}^{-1}=I$. Therefore the specified matrix, which is not a scalar identity matrix, is not diagonalizable.

For the displayed real symmetric matrix, choose the orthogonal matrixIts columns form an orthonormal eigenbasis, andIn the orthogonal coordinates $u=(x+y)/\sqrt{2}$, $v=(x-y)/\sqrt{2}$, the quadric surface has equation $u^2+3v^2+3z^2=1$. It is a prolate spheroid, with long semiaxis one in direction $(1,1,0)$ and two equal short semiaxes $1/\sqrt{3}$. The figure gives its principal-axis sketch.

A diagonalizable matrix can certainly have exactly two distinct eigenvalues: $\mathrm{diag}(1,1,2)$ is already diagonal and has an eigenbasis despite its repeated eigenvalue.

Squaring with the product law givesEquality to the identity matrix requires $2a=2c=0$ in ${\mathbb{F}}_5$. Since 2 has a multiplicative inverse modulo 5, this forces $a=c=0$. The middle condition then gives $2b=0$, hence $b=0$. ThereforeThis agrees with Lagrange's theorem: a group of odd order has no element of order two.

The subset specified by $b\ne 0$ is not a subgroup. It omits the identity element. It also fails closure: both $A(0,1,0)$ and $A(0,4,0)$ lie in it, but their product is $A(0,0,0)=I$, whose $b$ coordinate is zero.

The explicit matrices areDirect matrix multiplication givesThe reverse products have the opposite signs: $YX=\hat{Z}$, $ZY=\hat{X}$, $XZ=\hat{Y}$. Since $W$ is the identity matrix and $X^2=Y^2=Z^2=-W$,These are the usual quaternion multiplication relations in the complex matrix representation of quaternions.

Write a matrix in the set as $M(a,b)$. The zero matrix $M(0,0)$ has no multiplicative inverse matrix, so $S$ is not a group. Its determinant iswhich vanishes only at that zero matrix. It is therefore the only element that needs to be removed.

For closure, direct matrix multiplication givesThe product of nonzero determinants is nonzero, so two nonzero elements never multiply to zero. The identity matrix is $M(1,0)$, associativity is inherited from matrix multiplication, andThus $\boxed{S\setminus \{0_{2\times 2}\}\text{ is a group}}$. It is the multiplicative group of nonzero quaternions in the complex matrix representation of quaternions.

Choose an orthonormal basis with $\mathbf{n}={\mathbf{e}}_3$. A half-turn about $\mathbf{n}$ kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar identity matrix plus an antisymmetric matrix. The resulting oriented-axis invariant Cartesian tensor, in coordinate-free form, iswith real constants $a,d,h$. The Kronecker delta, the outer product $n_i{n}_j$, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about $\mathbf{n}$, so this form is also sufficient. In the chosen basis,The antisymmetric part must be retained unless further symmetry is imposed.

Decompose the electric field as $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=(\mathbf{n}\cdot \mathbf{E})\mathbf{n}$. The electric current density isFor a nonzero real electric field, the alignment of current and field in an axisymmetric conductor gives the complete classification of collinearity:

For completeness, if ${\mathbf{E}}_{\perp }\ne 0$ and $\mathbf{J}=\lambda \mathbf{E}$, taking the dot product with $\mathbf{n}\times {\mathbf{E}}_{\perp }$ forces $h=0$. The transverse equation then gives $\lambda =a$, while a nonzero axial component gives $\lambda =d$. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for $\mathbf{J}$, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.

The cyclic property of the trace impliesTaking the matrix trace of the asserted equality therefore gives $nc=0$. For a positive integer $n$ over the complex numbers this requires $\boxed{c=0}$. It is also sufficient: choose $M=N=0$, or any two commuting matrices. Thus a finite-dimensional operator commutator cannot be a nonzero scalar multiple of the identity matrix.

The disk still has only three degrees of freedom and hence three normal modes: two centre coordinates and one in-plane rotation. Five springs alter the restoring quadratic form but add no disk coordinates. The fivefold rotational symmetry makes the translational stiffness proportional to the two-dimensional identity matrix, and makes translation-rotation coupling vanish. Therefore the two translation modes are always degenerate; the rotational mode is separate, with possible accidental equality of its frequency for particular parameters. The general formulas in small oscillations of a disk attached to prestretched radial springs apply with $N=5$.

A Givens rotation acts as the identity matrix outside two selected coordinates $i,j$, and has the following block on those coordinates:The convention with the opposite signs of the sines is equivalent after replacing $\theta$ by $-\theta$. A Householder reflection with nonzero normal vector $w\in {\mathbb{R}}^n$ isIt reverses the component parallel to $w$ and fixes the perpendicular hyperplane. Both are orthogonal matrices, so their transposes are their inverses and they preserve the Euclidean norm.

A real square matrix $Q$ is an orthogonal matrix when $Q^{T}Q=I$, equivalently its columns form an orthonormal basis. Here $Q^T$ is the transpose and $I$ the identity matrix. For every vector $x$,Taking nonnegative square roots provesThus an orthogonal matrix preserves the Euclidean norm.

Let $r=\sqrt{x_i^2+x_j^2}$. If $r>0$, choose $\theta$ so thatSuch an angle exists because the two ratios have squared sum one. Then the changed coordinates areAll other coordinates satisfy $y_k=x_k$. If $r=0$, both input coordinates are zero; take $\theta =0$, giving the identity matrix on this pair. In all cases,This fixes a nonnegative retained coordinate and handles division by zero explicitly. For floating-point computation, the stable construction of a Givens rotation uses scaled Euclidean norm evaluation to avoid unnecessarily overflowing or underflowing the squares.