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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.

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.

Write $C=(c_{ij})$ and $(d_1,d_2,d_3)=(0,2,4)$. Because $D$ is a diagonal matrix,Thus $CD=DC$ exactly when $(d_j-d_i)c_{ij}=0$ for every $i,j$. The three eigenvalues are distinct, so every off-diagonal entry of $C$ must be zero; diagonal entries are unrestricted. Consequently the centralizer of a diagonal matrix here isEvery displayed matrix indeed commutes with $D$, since diagonal matrices multiply entry by entry on the diagonal.

The transpose is a linear map, so $T(A)=A+A^T$ is a linear map on the $n^2$-dimensional vector space $M_{n\times n}(F)$. Let $E_{ij}$ denote a matrix unit. The answer depends on the characteristic of a field.

If $\mathrm{char}F\ne 2$, then $T(A)=0$ means $A^T=-A$. Such skew-symmetric matrices have zero diagonal, and their entries above the diagonal can be chosen freely. A basis of the kernel of a linear map is $E_{ij}-E_{ji}$ for $i<j$, giving nullity $n(n-1)/2$. The image of a linear map consists of all symmetric matrices: $T(A)$ is symmetric, and every symmetric matrix $S$ is $T(S/2)$. Its basis comprises $E_{ii}$ and $E_{ij}+E_{ji}$ for $i<j$, giving rank of a linear map $n(n+1)/2$.

If $\mathrm{char}F=2$, then $-A=A$, so $\ker T$ consists of all symmetric matrices, with dimension $n(n+1)/2$. Every $T(A)$ is symmetric with zero diagonal, and every symmetric zero-diagonal matrix $S$ arises by taking $A_{ij}=S_{ij}$ for $i<j$ and all other entries zero. The image of a linear map therefore has basis $E_{ij}+E_{ji}$ for $i<j$, and dimension $n(n-1)/2$. This is the symmetrization map in characteristic two.

In both cases the rank-nullity theorem is verified explicitly:

Perform the elementary row operations $R_2\leftarrow R_2+2R_1$ and $R_3\leftarrow R_3-R_1$, then $R_2\leftarrow R_2/2$, $R_1\leftarrow R_1-R_2$, and finally $R_1\leftarrow R_1+R_3$, $R_2\leftarrow R_2-3R_3$:Consequently the reduced row echelon form is $I_3$. Applying the same operations to the augmented column $b=(b_1,b_2,b_3{)}^T$ givesThus every $b\in {\mathbb{R}}^3$ admits exactly one solution. This follows because the elementary row operations preserve the solution set of the augmented linear system, and the reduced coefficient matrix is the identity 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.

We use the monic polynomial convention for the characteristic polynomial of an $n\times n$ matrix:It has degree $n$, since the only contribution to the coefficient of $t^n$ in the determinant is the product of the $n$ diagonal terms $t$. The alternative convention $\det (A-t{I}_n)=(-1{)}^n{\chi }_A(t)$ has the same polynomial roots.

The matrix units $E_{ij}$, $1\le i,j\le n$, form a basis of $M_n(\mathbb{R})$: each matrix has the unique expression $A={\sum }_{i,j}{a}_{ij}{E}_{ij}$. There are $n^2$ of them, so

Yes. Consider the following $n^2$ matrices:Each has only row $i$ nonzero and has $i$th diagonal entry $1$, so part (c)(ii) proves that it is a diagonalisable matrix. The collection is a spanning set, because it contains every $E_{ii}$ andThus it spans the matrix unit basis. A spanning set of $n^2$ vectors in an $n^2$-dimensional vector space is a basis. Alternatively, in a vanishing linear combination of $\mathcal{B}$, each off-diagonal entry first forces its own coefficient to be zero, and then each diagonal entry forces the remaining coefficient to be 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 the Einstein summation convention, with repeated component indices summed. The following three expressions are listed separately below. For cyclic invariance of the trace,where the middle equality commutes scalar entries, not the matrices themselves. If $A$ is invertible, this givesThis is similarity invariance of trace.

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

Subtract corresponding entries in the two matrices:Both matrices have shape $2\times 3$, so this subtraction is defined.

The homogeneous linear system has coefficient matrixNonzero solutions exist exactly when its determinant vanishes. Expanding along the first row givesThus $\lambda =1$ or $-6$. The first equation always gives $x=3z$. For $\lambda =1$, the second then gives $y=-2z$; the third imposes the same relation. For $\lambda =-6$, the second gives $y=3z/2$, again consistent with the third. Therefore the nonzero solutions areFor every other $\lambda$, the matrix is invertible and only the zero solution exists.

Here the vertical bars denote determinants. Since $B=B^{-1}$, taking determinants gives $(\det B{)}^2=1$. The condition $\det B>0$ selects $\det B=1$. By the multiplicative property of the determinant,The hypotheses determine this determinant, while allowing many possible matrices $A$ and $B$.

The required matrix form of the linear system isSubtract the first row from the third and expand the determinant along the resulting row:Thus the linear system has a unique solution, for every real $\mu$, precisely when $\boxed{\lambda \ne 1,2}$.

For $\lambda =0$, the second equation gives $x=\mu$. Subtracting the first equation from the third gives $-y=1$, and the first equation then determines $z$:For $\lambda =1$, the first and third left-hand sides are identical but their right-hand sides are $1$ and $2$. HenceFor $\lambda =2$, subtracting the first equation from the third gives $y=1$. The first equation now gives $x+2z=0$, while the second requires $x+2z=\mu$. ConsequentlyThe free variable in a linear system in the consistent singular case describes a line of solutions. These direct reductions distinguish an inconsistent linear system from one with infinitely many solutions, which a zero determinant alone cannot do.

Call the given matrix $M$. Its middle coordinate decouples, immediately giving the eigenvalue $3$ with normalized eigenvector $(0,1,0{)}^T$. The remaining characteristic polynomial comes from the $xz$ block:For $\lambda =-4$, the first row of $(M+4I)v=0$ gives $x=\sqrt{3}z$; for $\lambda =8$, the first row of $(M-8I)v=0$ gives $z=-\sqrt{3}x$. After vector normalization, one possible complete set isEach has unit Euclidean norm. The dot products with $v_2$ vanish because the other two middle components are zero, and $v_1\cdot v_3=(-\sqrt{3}+\sqrt{3})/4=0$. This explicitly verifies orthonormality, also predicted by the spectral theorem for real symmetric matrices with distinct eigenvalues. Changing any eigenvector's sign gives an equally valid normalized choice.

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.