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For square matrices $A,B$ over a field,If $A$ is invertible, the products are similar matrices. In general, apply that case to $(A-sI)B$ and $B(A-sI)$ for all but finitely many $s$, then use polynomial identity in $s$.

Under an orthogonal change of coordinates represented by $Q$, the component matrix changes byThus $T^{\prime }$ is similar to $T$, so its characteristic polynomial, eigenvalues, and their algebraic multiplicities are unchanged.

If $T$ is symmetric, the spectral theorem for real symmetric matrices makes it diagonalizable, so algebraic and geometric multiplicities agree. Consequently the dimensions of its eigenspaces are also independent of the coordinate frame.

For the stated tensor, choose a unit eigenvector $n$ belonging to the simple eigenvalue $\mu$. Its perpendicular plane is the eigenspace of $\lambda$, soHence

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

Two square matrices are similar when $B=P^{-1}AP$ for some invertible matrix $P$. A Jordan normal form is a block-diagonal matrix whose blocks have one eigenvalue on the diagonal, ones on the superdiagonal, and zeros elsewhere; over an algebraically closed field every matrix is similar to such a form, unique up to reordering its Jordan blocks.

Both displayed matrices have characteristic polynomialHowever,so their eigenspace dimensions are respectively two and one. Thus $A$ has Jordan form $J_2(1)\oplus J_1(1)$, whereas $B$ has Jordan form $J_3(1)$. By uniqueness of Jordan normal form,

Given a solution $u(t)$, construct $a_j(t)=\frac{1}{2}{e}^{u_j(t)/2}$ and hence the smooth matrices $L(t)$ and $B(t)$. Let $U(t)$ solve the linear ordinary differential equationBecause $B(t)$ is skew-symmetric,so $U(t)$ is an orthogonal matrix and is invertible. Moreover,It follows thatThus the Lax pair flow is an isospectral Lax equation: every $L(t)$ is similar to $L(0)$ and has the same eigenvalues and characteristic polynomial.

Every symmetric function of these constant eigenvalues is a first integral. In particular, the trace invariants of a Lax equationare constant because the cyclic property of the trace givesFor example,is an explicit first integral of equation $(†)$.