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There aresuch conjugacy classes. Every matrix in $G{L}_{10}(\mathbb{Q})$ is an invertible matrix, so by the minimal polynomial of an invertible matrix its minimal polynomial must have nonzero constant term. Buthas constant term zero and cannot be the minimal polynomial of an invertible matrix.

The rank of a matrix is the dimension of its column space, equivalently the dimension of the image of a linear map represented by the matrix. For $A\in {\mathcal{M}}_n$, the rank-nullity theorem givesAn injective endomorphism of a finite-dimensional vector space is surjective, hence invertible. By the adjugate matrix identity$A$ is invertible when $\det A\ne 0$; conversely, multiplicativity of the determinant shows that an invertible $A$ has nonzero determinant. Therefore

Let $E_{ij}$ be the matrix unit with its only nonzero entry at $(i,j)$. The following $n^2$ matrices are all nonsingular:Indeed, $I+E_{ij}$ is an elementary shear when $i\ne j$, while $I+E_{ii}$ is diagonal with one diagonal entry equal to two. Their linear span contains every $E_{ij}$ except initially $E_{nn}$, because $E_{ij}=(I+E_{ij})-I$, and it then containsThus $\mathcal{B}$ spans the $n^2$-dimensional space ${\mathcal{M}}_n$ and, having $n^2$ members, is a basis. This also covers $n=1$, when $\mathcal{B}=\{I\}$.

Now let $A$ be a nonsingular zero-one matrix. If it had fewer than $n-1$ zero entries, at least two rows would contain no zero at all. Those two rows would both be the all-one row, contradicting linear independence. Hence every such matrix has at mostones. The bound is attained. Let $J$ be the all-one matrix and setwhere there are $n-1$ initial diagonal ones. If $A\mathbf{x}=0$ and $S={\sum }_j{x}_j$, the first $n-1$ row equations give $x_i=S$, while the last gives $S=0$. Hence every $x_i=0$, so $A$ is nonsingular and has exactly $n^2-n+1$ ones. Therefore

Fix $I$ and write $I^{\prime }$ for the complementary column indices. Over $U_I$, decompose $M=(M_I,M_{I^{\prime }})$ and $v=(v_I,v_{I^{\prime }})$. The equation $Mv=0$ is equivalent toMatrix inversion is smooth on the invertible matrices, sois a smooth, fiberwise-linear trivialization $U_I\times {\mathbb{R}}^{n-m}\to {\pi }^{-1}(U_I)$. The sets $U_I$ cover the base, proving that $\pi :E_{m,n}\to X_{m,n}$ is a vector bundle of rank $n-m$. It is the kernel bundle of a constant-rank family of linear maps, and its rank also follows from the rank-nullity theorem.

An elementary row operation is one of three reversible changes: interchange two rows; multiply one row by a nonzero scalar; or add a scalar multiple of one row to a different row. Their inverses are, respectively, the same interchange, multiplication by the reciprocal, and addition of the negative scalar multiple.

Applied to an augmented matrix, these changes interchange linear equations, rescale an equation by a nonzero factor, or add a multiple of one equation to another. Every old solution satisfies the new equations. Applying the inverse elementary row operation gives the converse, so the solution set is unchanged at every step. Equivalently, all the operations together givewhere $E$ is an invertible matrix, a product of elementary matrices.

A system of linear equations $T_{\alpha }x=b$ is solvable exactly when $b\in \mathrm{im}{T}_{\alpha }$. If $x_0$ is one solution, thenso its solution set is $x_0+\ker T_{\alpha }$, a translate of the kernel of a linear map.

For $\alpha \ne 2$, the matrix is an invertible matrix, andEvery $b$ is in the image of a linear map and has exactly one preimage. For $\alpha =2$, its first two columns span the image of a linear map, while the corresponding homogeneous linear system gives the kernel of a linear map:The prescribed $b=(-1,\beta ,1{)}^T$ satisfies the displayed condition precisely when $\beta =0$. If $\beta \ne 0$, it lies outside the image of a linear map and there is no solution. If $\beta =0$, it lies in the image of a linear map, and the nonzero kernel of a linear map supplies infinitely many solutions.

Choose an ordered basis $\mathcal{B}$ of $V$ and let $A=[T{]}_{\mathcal{B}}$. Define the characteristic polynomial byIf a different basis gives $B=P^{-1}AP$ for an invertible matrix $P$, multiplicativity of the determinant givesThus the definition is independent of the chosen basis. The alternate convention $\det (A-xI)$ differs by the constant factor $(-1{)}^{\dim V}$ and gives the same roots.

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.

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.

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 an invertible matrix, thenIf instead $B$ is an invertible matrix, thenThus in either case $AB$ and $BA$ are similar matrices. Explicitly, whenever $Y=P^{-1}XP$ and $Xu=\lambda u$ with $u\ne 0$, one has $Y(P^{-1}u)=\lambda P^{-1}u$ and $P^{-1}u\ne 0$. Applying the same argument with $P^{-1}$ in place of $P$ proves equality of the sets of eigenvalues, including zero.

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.

Use $A=M$ and $B=N$ from part (a). The proposed unreversed product iswhich differs from the actual inverse $\left(\begin{array}{cc}-25& 18\\ 7& -5\end{array}\right)$. Both original matrices are invertible matrices, so this is a valid counterexample. The universally valid order is $\boxed{(AB{)}^{-1}=B^{-1}{A}^{-1}}$.

Take $A=B=I_2$, which are invertible matrices. Then $\det (A+B)=\det (2I_2)=4$, whereas $\det A+\det B=2$. This proves that the determinant is not additive. Being a multilinear map of separate columns does not imply additivity in the entire matrix.

A two-sided matrix inverse requires a square matrix, so $\boxed{n=3}$ is necessary. It exists only if that square matrix is also an invertible matrix, equivalently $\det M\ne 0$; choosing the dimension alone does not ensure invertibility.

False. The cyclic property of the trace permits $ABC\mapsto BCA$, but not an arbitrary interchange. Even with $A$ and $B$ both invertible matrices, takeThen $\mathrm{Tr}(ABC)=1$ and $\mathrm{Tr}(BAC)=2$, providing a counterexample.

Write the linear system as $Ar=d$ withExpansion along the first row gives $\det A=(16-15)-2(12-5)+3(9-4)=2$. Since $\det A=2\ne 0$, $A$ is an invertible matrix, and $r=A^{-1}d$ is the unique solution for this right side.

For the two-by-two matrix, take the product of the diagonal entries minus the product of the off-diagonal entries:In particular, $A$ is an invertible matrix.

The coefficient matrix has determinant $-1$, so it is an invertible matrix. Its homogeneous linear system therefore has only the zero solution:One can also subtract the first equation from the second to get $z=0$, then subtract the first from the third to get $y+2z=0$, giving $y=0$ and finally $x=0$. Thus there are no nonzero solutions.

Suppose $B\mathbf{x}=0$. The given quadratic form bound then givesso $\mathbf{x}=0$. Thus the kernel of $B$ is trivial. A square matrix with trivial kernel is an invertible matrix, proving the assertion. No symmetry assumption on $B$ is needed.