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A linear endomorphism is diagonalizable over its field precisely when its minimal polynomial is a product of distinct linear polynomials over that same field. Splitting alone is insufficient: repeated factors permit nontrivial Jordan blocks.

For the displayed matrix, a determinant expansion givesThese formulas can be reduced in each requested characteristic. The four cases below distinguish failure to split from failure to have distinct factors.

The proposed equivalence is false. On a nonzero vector space over any field of characteristic different from two, chooseThese linear projections commute and satisfy $EFG=0$. Nevertheless their sum is $2I$, andThus the proposed conditions are not sufficient. For commuting linear projections the actual obstruction is the sum of the pairwise products:The single condition on the triple product does not remove this obstruction.

Both kernels are integrable functions, so the convolution theorem givesThe uniqueness of the Fourier transform on integrable functions gives equality almost everywhere. Both sides are continuous, so it holds at every point:For completeness, continuity of the convolution follows by dominated convergence: $q_t$ is bounded and continuous, so for $x_j\to x$ the integrands $q_t(x_j-y)q_s(y)$ converge pointwise and are dominated by $\Vert q_t{\Vert }_{\infty }{q}_s(y)$.

The identity is the heat semigroup property: evolving for time $s$, then for time $t$, is the same as evolving once for time $s+t$.

Complete the square inside the Fourier transform of the Gaussian heat kernel:After $z=x/(2\sqrt{t})$,The supplied shifted Gaussian integral equals $\sqrt{\pi }$ for real $\sqrt{t}\omega$. HenceAt zero this says $\int q_t=1$, so convolution with the heat kernel averages rather than rescales the initial mass. The formula is an instance of the Fourier transform of a Gaussian.

The profile is an even function, so its Fourier transform is real andFor $\omega \ne 0$, integration by parts givesThereforeAt zero, direct integration gives the area of the triangle, namely two; the displayed quotient has the same limit. In particular $\hat{T}(\omega )\ge 0$ for every real frequency. This is the Fourier transform of a triangular function.

One can also see the nonnegative square structurally: $T=\frac{1}{2}{1}_{[-1,1]}\ast 1_{[-1,1]}$, so the convolution theorem gives $\hat{T}=\frac{1}{2}(2\sin \omega /\omega {)}^2$.

Use the Fourier transform conventionThe absolute integrability of $v$ makes this integral exist for every real $\omega$, with $|\hat{v}(\omega )|\le \Vert v{\Vert }_1$. With this convention the inverse Fourier transform has factor $1/(2\pi )$; maintaining that normalization is essential in part (e).

For a real solution, $u^4\ge 0$, so $\Delta u\ge 0$. The maximum principle for subharmonic functions gives $u\le {\max }_{\partial D}u=0$. The additional nonnegativity condition gives $u\ge 0$. ThereforeThis function satisfies both the equation and boundary condition, so it is exactly the full set of nonnegative solutions. The sign restriction is used alongside the maximum principle, rather than treating the nonlinear equation as a linear uniqueness problem.