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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:

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.

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.

Assume $A\ne 0$, as is implicit in the claimed conclusion. Suppose that a surjective ring homomorphism $f:R\to M_2(A)$ exists. Choose $r_1,r_2\in R$ mapping to the matrix units $E_{12}$ and $E_{21}$. Applying $f$ to the assumed relation gives integers $c_1,\dots ,c_n$ such thatBut $E_{12}^2=0$ and $E_{12}{E}_{21}=E_{11}$, so the right side is $c_1{E}_{11}$. Its $(2,2)$ entry is zero, whereas the $(2,2)$ entry of $E_{22}$ is $1_A\ne 0$. This contradiction proves that no such surjective ring homomorphism exists.

If the convention for rings allows $A$ to be the zero ring, that case must be excluded from the assertion: $M_2(0)=0$, and the zero ring homomorphism is surjective. This is a convention issue in the printed question.

Let $e_{ij}$ denote the matrix unit with one at position $(i,j)$ and zero elsewhere. The five inverses areThus each is the identity matrix with only the indicated off-diagonal matrix entry altered. These inverses reverse the corresponding row additions; no row scaling or permutation is involved.

The zero matrix has the required pattern, and matrix addition and scalar multiplication preserve every prescribed zero entry. This proves the vector subspace condition. A basis consists of the matrix unitsThey independently select every permitted entry of a tridiagonal matrix, so the dimension is $n+2(n-1)=3n-2$ for $n\ge 1$.

Let $E_{ij}$ denote a matrix unit, with its only nonzero entry equal to one in row $i$, column $j$. A basis for the vector space of real skew-symmetric matrices isEach upper-triangular entry is free, its reflected lower-triangular entry is its negative, and every diagonal entry is zero. This is the basis of the space of real skew-symmetric matrices.

A simple unital associative algebra is nonzero and has no nonzero proper two-sided ideals. Finite-dimensional complex examples are exactly the full matrix algebras $M_d(\mathbb{C})$, by the Artin–Wedderburn theorem. A simple associative algebra need not be a division algebra: for $d>1$ it has noninvertible nonzero matrices. Any nonzero two-sided ideal of $M_d(\mathbb{C})$ contains a matrix unit, by multiplying a nonzero entry on the left and right by suitable matrix units, and therefore contains all of them.

For the symplectic Lie algebra $\mathfrak{sp}(4)$ preserving $J=\left(\begin{array}{cc}0& I_2\\ -I_2& 0\end{array}\right)$, write $h_1=E_{11}-E_{33}$ and $h_2=E_{22}-E_{44}$. Its positive roots of a root system have sl2 subalgebra associated with a root triplesThe matrix unit commutator identity verifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ and the two eigenvalue relations $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$, $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$.

Over $\mathbb{R}$ or $\mathbb{C}$, every linear functional on all $n\times n$ matrices satisfying $f(AB)=f(BA)$ is a scalar multiple of the trace. The matrix units show that the functional vanishes off the diagonal and has equal values on every diagonal unit. The normalization $f(I)=n$ fixes the scalar to one.

Every commutator of square matrices is traceless by cyclicity of the trace. Conversely, off-diagonal matrix units and differences of diagonal units are themselves commutators and span all traceless matrices. This proves equality with the linear span of commutators, without needing a theorem about representation as one commutator.