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The characteristic polynomial ${\chi }_A(t)=\det (tI-A)$ has degree $n$. By the fundamental theorem of algebra, it has a complex root $\lambda$. Henceso $\lambda I-A$ has a nonzero kernel. Any nonzero $v$ in that kernel satisfiesThus $\lambda$ is an eigenvalue and $v$ is a corresponding eigenvector.

Eigenvectors belonging to distinct eigenvalues are linearly independent. Indeed, supposeis a shortest nontrivial relation, where $A{v}_j={\lambda }_j{v}_j$ and the ${\lambda }_j$ are distinct. Applying $A-{\lambda }_{m}I$ givesa shorter nontrivial relation, a contradiction.

Thus $n$ distinct eigenvalues provide a basis of $n$ eigenvectors, proving that $A$ is diagonalizable over $\mathbb{C}$. If $A$ is real and all its eigenvalues are real, each real singular matrix $A-\lambda I$ has a nonzero real null vector. Choosing one for each eigenvalue gives a real eigenbasis. Hence the real matrix is diagonalizable over $\mathbb{R}$. This is the distinct eigenvalues imply diagonalizability criterion.

Write the complex similarity matrix aswith $X,Y$ real. From $C=SD{S}^{-1}$ we obtain $CS=SD$, and equating real and imaginary parts givesThereforefor every real $t$.

Now $p(t)=\det (X+tY)$ is a real polynomial andThus some real $t$ has $p(t)\ne 0$. Taking $T=X+tY$, we have a real invertible matrix satisfying $CT=TD$, orThis proves real similarity from complex similarity.

Consider the real polynomialThe hypothesis says $p(i)\ne 0$, so $p$ is not identically zero. A nonzero polynomial has only finitely many roots, whereas $\mathbb{R}$ is infinite. Hence there is a real $\lambda$ withThus $A+\lambda B$ is invertible, as summarized by the real parameter avoiding a singular matrix pencil.

The characteristic polynomial of $A$ isA scalar $\lambda \in \mathbb{C}$ is an eigenvalue when there is a nonzero vector $v$ such thatSuch a $v$ is an eigenvector belonging to $\lambda$. Equivalently, $\lambda$ is a root of ${\chi }_A$ and the corresponding eigenvectors are the nonzero elements of $\ker (A-\lambda I)$.

The matrix trace isUsing matrix multiplication and interchanging dummy indices,

The proposed noncyclic interchange is false. With matrix units,one hasTrace is invariant under cyclic permutations, but an arbitrary swap need not be cyclic.

For $a=(a_1,a_2,a_3{)}^T$, the cross-product matrix isIts diagonal is zero, so $\mathrm{tr}A=0$. Also $Aa=a\times a=0$, and $a\ne 0$, so $A$ has a nontrivial kernel. Hence

Write $a=(a_1,a_2,a_3,a_4{)}^T$. The reflection in a hyperplane orthogonal to $a$ sendssoIt acts as $-1$ on $\mathrm{span}\{a\}$ and as $1$ on the three-dimensional hyperplane $a^{\perp }$. Therefore

Since $P=P^{\prime }=0$, it suffices to map the two edge vectors. Let the columns of the required matrix be $u_1,u_2,u_3$. The condition $MQ=Q^{\prime }$ givesThe condition $MR=R^{\prime }$ then givesand henceThese are orthonormal. Taking $u_3=u_1\times u_2=(0,-1,0{)}^T$ produces the orthogonal matrixIts columns are orthonormal, and direct substitution verifies $MP=P^{\prime }$, $MQ=Q^{\prime }$, and $MR=R^{\prime }$.

PutSince $x,y,z$ are unit vectors,The feasible inner products with two unit vectors satisfyThe desired pair $p=q=1/2$ is feasible exactly when $c\ge -1/2$. In that case every maximizer has these inner products. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point isUsing the vector triple product,Hence

Rotations preserve dot products and cross-product lengths, while $z^{\prime }=2z$. ThusLetThe target pair is feasible precisely whenorOn this interval,

Outside this interval, let $\lambda >0$ be the unique number for whichand write $r=(p,q{)}^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing inner products solely from $c=y\cdot z$. In the remaining cases,Neither answer depends on ${\theta }_1$ or ${\theta }_2$.

By the Cauchy-Hadamard theorem, putThe squared coefficients have root limsup $L^2$, so their radius is $R^2$. Equal finite nonzero radii therefore require $R=R^2$, hence $R=1$. This can happen: for $a_n=n+1$, both $\sum a_n{z}^n$ and $\sum a_n^2{z}^n$ have radius one.

For the coefficients $a_n^{a_n}$, the relevant root terms areEquality can again occur. For example, takeThen $a_n\to \infty$, $a_n^{1/n}\to 1$, andThus both $\sum a_n{z}^n$ and $\sum a_n^{a_n}{z}^n$ have radius one. These examples illustrate the radius of convergence after powering coefficients.

Suppose the series converges at $w\ne 0$. Then its terms $a_n{w}^n$ are bounded, say by $M$. Whenever $|z|<|w|$,so comparison with a geometric series proves absolute convergence at $z$.

Letwith the natural values $0$ or $\infty$ allowed. If $|z|<R$, choose a convergent point $w$ with $|z|<|w|\le R$; the preceding argument gives convergence at $z$. If $|z|>R$, convergence at $z$ would contradict the definition of $R$. Hence the series converges for $|z|<R$ and diverges for $|z|>R$. No universal assertion is possible on $|z|=R$.

Another application of the Cauchy-Schwarz inequality givesWhen $p>1/2$, the second factor is bounded because $2p>1$, so the series converges.

At the endpoint, take $a_1=1$ and, for $n\ge 2$,Then $\sum a_n$ converges, whilewhose series diverges. This proves both parts of weighted square roots of a summable sequence.