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If $A$ is an invertible matrix, thenThus $AB$ and $BA$ are similar matrices, and similar matrices have the same characteristic polynomial.

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.

A square matrix is singular when it is not invertible, equivalently when its determinant is zero or its kernel is nontrivial.