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The definition of matrix multiplication givesThe middle equality uses commutativity of scalar multiplication in $\mathbb{R}$ or $\mathbb{C}$, together with interchange of finite sums. Thus the trace is unchanged by cyclically moving the two factors.

Give $V_n$ the action $(g\cdot P)(x,y)=P((x,y)g)$, where $(x,y)$ is a row vector. Substitution preserves homogeneous degree, and composition agrees with matrix multiplication, so this is a complex representation of SU(2) group. The diagonal torus $\mathrm{diag}(z,z^{-1})$ acts on $x^{n-j}{y}^j$ with weight $z^{n-2j}$; these weights are distinct.

A nonzero invariant subspace is invariant under the torus, so finite-dimensional Fourier projection onto its weight spaces supplies a nonzero monomial. Differentiating the group action and complexifying supplies the operators $E=x{\partial }_y$ and $F=y{\partial }_x$. Their repeated application moves between every successive monomial, with nonzero coefficients until an endpoint is reached:The invariant subspace therefore contains the whole monomial basis. ThusFor $n=0$ this is the one-dimensional trivial representation. Torus weight projections are legitimate because an invariant complex subspace is closed in finite dimension.

For a nonzero vector $v\in {\mathbb{R}}^n$, a Householder reflection isIt is immediately symmetric. If $P=v{v}^T/(v^{T}v)$, then $P^2=P$, soHence $H$ is also an orthogonal matrix, with $H^{-1}=H$.

The similarity transformation can be expanded asComputing $v^{T}A$, $Av$, and the scalar $v^{T}Av$ costs $O(n^2)$ arithmetic operations, after which the remaining updates are outer products and scalar multiples, also costing $O(n^2)$. Thus $HA{H}^{-1}$ can be formed in $O(n^2)$ operations rather than by two general $O(n^3)$ matrix multiplications.

If $M_1^{T}J{M}_1=J$ and $M_2^{T}J{M}_2=J$, then matrix multiplication and the transpose rule giveThus the product also satisfies the equation.

The equation and part (i) imply that $M_1$ is invertible. Multiplying
$M_1^{T}J{M}_1=J$ on the left by $(M_1^{-1}{)}^T$ and on the right by $M_1^{-1}$ givesHence the set of solutions is closed under products and matrix inverses.

Write an element of the Integer Heisenberg group as $(x,y,z)$. Matrix multiplication givesThe subgroupis normal and isomorphic to ${\mathbb{Z}}^2$. If $t=(1,0,0)$, thenThus, on the coordinate column $(y,z{)}^T$, conjugation by $t$ is the linear map with matrixEvery element has a unique expression $(0,y,z)t^x$, so

The commutators $[t,(0,y,z)]$ fill the central subgroup of matrices $(0,0,z)$, while the quotient by this subgroup is generated freely and abelianly by the images of $(1,0,0)$ and $(0,1,0)$. Equivalently, $(A-I){\mathbb{Z}}^2$ is the second coordinate axis. Therefore the abelianization is

For any matrix $Y$ and compatible invertible matrix $P$, the linear map $r\mapsto rP$ maps $\mathrm{rowspace}(Y)$ onto $\mathrm{rowspace}(YP)$: a linear combination of rows of $Y$ is sent to the same linear combination of rows of $YP$. Its inverse function is $s\mapsto s{P}^{-1}$, so these row spaces have equal dimensions. Hence right matrix multiplication by $P$ also preserves row rank.

If $X=P^{-1}YP$, ii now givesThis proves the required matrix similarity invariant without needing to assume equality of row rank and column rank.

When $A$ is an invertible matrix, $B=A^{-1}(AB)$. Apply i first to $AB$ and then to $A^{-1}(AB)$:Thus the two row spaces are equal. In particular, invertible left matrix multiplication preserves row rank.

Write $A=(a_{ij})$ as an $m\times r$ matrix and $B$ as an $r\times n$ matrix, with rows $b_1,\dots ,b_r$. The rule for matrix multiplication givesEvery row of $AB$ is therefore a linear combination of rows of $B$. Taking linear spans gives

Matrix similarity transports commuting operators: with $M=PD{P}^{-1}$,By ii, all such matrices are thereforeFor completeness, matrix multiplication givesThe parameterization is injective, since matrix multiplication by $P^{-1}$ and $P$ recovers the parameter diagonal matrix. Thus there are infinitely many choices. Already the distinct scalar matrices $t{I}_3$, $t\in \mathbb{R}$, give an infinite family.

For the given real matrix $M$, direct matrix multiplication gives the eigenvectorsPut these eigenvectors into the columns ofSince $\det P=-2\ne 0$, $P$ is an invertible matrix. The three column identities give $MP=PD$, soIn particular, the matrix similarity is established by an explicit change-of-basis matrix.

Bilinearity of the real inner product gives, for the specified linear combinations,The last equality is the entrywise rule for matrix multiplication, with $x^T$ the transpose of the column vector $x$.

First find the fixed axis of rotation by solving $(M-I)\mathbf{n}=0$. After removing a nonzero common factor, the linear equations areThey give $n_2=n_3=-n_1$. Choose the following unit vectors:Their dot products show that they form an orthonormal basis; hence $P=({\mathbf{p}}_1{\mathbf{p}}_2{\mathbf{p}}_3)$ is an orthogonal matrix. Since $M$ is an orthogonal matrix fixing ${\mathbf{p}}_1$, it preserves the orthogonal complement of ${\mathbf{p}}_1$. Its restriction there has determinant $1$, so it is a planar rotation.

The trace gives $1+2\cos \theta =\mathrm{tr}M=9/7$, so $\cos \theta =1/7$. To determine the sign of the sine, rather than just its magnitude, calculateThus a suitable angle and change of basis areIn fact direct matrix multiplication gives

For an $m\times n$ matrix, the corresponding elementary matrices are $m\times m$, because the operation is implemented by left matrix multiplication. Apply the desired elementary row operation to the identity matrix $I_m$. Writing $E_{ij}$ for the matrix unit with its only nonzero entry, $1$, in position $(i,j)$ gives:Indeed, these matrix products interchange rows $i,j$, scale row $i$, and add $c$ times row $j$ to row $i$, respectively. The displayed inverses undo precisely those operations. Algebraically, $E_{ij}^2=0$ when $i\ne j$, so $(I_m+c{E}_{ij})(I_m-c{E}_{ij})=I_m$; the other two inverse identities follow immediately from their row actions. Thus every elementary matrix is an invertible matrix.

If $A$ is a strictly upper triangular matrix, then $A_{ij}=0$ unless $j-i\ge 1$. We prove by mathematical induction thatThe assertion for $k=1$ is the definition. In the matrix multiplication formulaa nonzero summand would require $\ell -i\ge k$ and $j-\ell \ge 1$, hence $j-i\ge k+1$. This proves the induction step. Since $j-i\le n-1$ for all entries, $\boxed{A^n=0}$. In particular, every strictly upper triangular matrix is a nilpotent matrix.

Take the nonzero vector $1=(1,\dots ,1{)}^T$. Its $i$th image entry under matrix multiplication is the sum of row $i$:Thus $A1=\lambda 1$, exhibiting $1$ as an eigenvector with eigenvalue $\boxed{\lambda }$.

The identity is the identity matrixLeft or right matrix multiplication by it preserves every entry, as is also immediate from the product formula above.

Parametrize the matrices byDirect matrix multiplication givesThis proves closure. The identity matrix is $h(0,0,0)$, and the formulagives both a left and a right inverse, as substitution in the product formula verifies. Thus each element is an invertible matrix, and the subgroup test shows that $H$ is a subgroup of ${\mathrm{GL}}_3({\mathbb{Z}}_3)$. It is the unitriangular group ${\mathrm{UT}}_3({\mathbb{F}}_3)$ and has $3^3=27$ elements.

Apply row-column matrix multiplication twice:The indices $i,j$ are free, while $k,\ell$ are summed.

If $Av=\lambda v$, then $A^{2}v={\lambda }^{2}v$. Alternatively, direct matrix multiplication givesThus the normalized eigenvectors $v_{+},v_{-}$ from part (b) have respective eigenvaluesIf $\sin 2\theta =0$, then $B=I$, and the whole two-dimensional space is its eigenspace. This degeneracy includes cases where the two eigenvalues of $A$ were distinct but opposite.

Using row-column matrix multiplication, the second row of the first matrix is $(1,6)$. Therefore