Normal matrix

A matrix is normal when $A{A}^{†}=A^{†}A$.

For two normal matrices of the same size, a pairing of their eigenvalues has squared Euclidean discrepancy at most the squared Frobenius norm of their difference. For Hermitian matrices, the increasing real eigenvalue order gives such a pairing, because it minimizes squared discrepancy. For real symmetric matrices, the norm square is $\mathrm{Tr}(A-B{)}^2$.

For equally sized Hermitian matrices and a real test function with Lipschitz bound one, the difference of its averages under their empirical spectral measures is at most $N^{-1/2}$ times the Frobenius norm of their difference. Pair the increasing eigenvalues, apply the triangle inequality and Cauchy-Schwarz inequality, then use the Hoffman–Wielandt inequality.

If a normal matrix satisfies $Ax=\lambda x$, then $A^{†}x=\overline{\lambda }x$. Indeed, $M=A-\lambda I$ is normal and $\Vert Mx{\Vert }^2=x^{†}{M}^{†}Mx=x^{†}M{M}^{†}x=\Vert M^{†}x{\Vert }^2$. Since $Mx=0$, also $M^{†}x=0$. This identity proves orthogonality of distinct eigenspaces and gives spectral restrictions for Hermitian matrices, skew-Hermitian matrices and unitary matrices.

For $A=\left(\begin{array}{cc}-i{\omega }_1& a\\ c& -i({\omega }_1-b)\end{array}\right)$ with real parameters,Thus the displayed conditions are necessary and sufficient for a normal matrix. In the family $a=1$, $b=c=q$, normality requires $q=-1$, not $q=1$. If $q=10/Re$, this is the formal value $Re=-10$ and no positive Reynolds number qualifies. At that value $A$ is a skew-Hermitian matrix and its exponential is unitary, so every initial direction has optimal energy amplification of a linear system equal to one.

For every finite-dimensional normal matrix $A$, the matrix 2-norm equals the spectral radius:This follows by unitary diagonalization of a normal matrix and unitary invariance of the matrix 2-norm.

A matrix is non-normal when it does not commute with its adjoint. Its eigenvectors need not be orthogonal, so combinations of individually decaying modes may grow temporarily.

Transient growth is finite-time amplification in a stable linear system. For a non-normal generator it can occur even when every eigenvalue has negative real part, because nonorthogonal eigenmodes can interfere constructively.

For $x^{\prime }=Lx$ and $E=\Vert x{\Vert }_2^2$, the Hermitian matrix $H=(L+L^{\ast })/2$ gives $\dot{E}=2x^{\ast }Hx$. Thus some initial condition grows immediately iff the largest eigenvalue of $H$ is positive. All trajectories have nonincreasing energy iff $H$ is a negative semidefinite matrix. This differs from requiring negative real parts for the eigenvalues of $L$: a non-normal matrix can satisfy modal decay while allowing transient growth.

For the matrix exponential $A(t)=e^{Lt}$, maximizing the energy ratio over nonzero initial conditions givesThe Rayleigh quotient achieves its maximum at an eigenvector of $A^{\ast }A$ with largest eigenvalue, equivalently a right singular vector of $A$. Such an optimal disturbance is generally not an eigenvector of $L$.

For $L=\left(\begin{array}{cc}{\lambda }_1& 0\\ 1& {\lambda }_2\end{array}\right)$ with distinct real negative eigenvalues, the matrix exponential is $A=\left(\begin{array}{cc}a& 0\\ b& d\end{array}\right)$, where $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The optimal energy amplification of a linear system isIts eigenvectors are nonorthogonal, and immediate energy growth occurs exactly when ${\lambda }_1{\lambda }_2<1/4$. If ${\lambda }_2>{\lambda }_1$, putting $\alpha =({\lambda }_2-{\lambda }_1{)}^{-1}$ gives $G(t)\sim (1+{\alpha }^2)e^{2{\lambda }_{2}t}$. The optimal asymptotic initial direction $(\alpha ,1)$ is an adjoint eigenvector, while the eventual state aligns with the right eigenvector $(0,1)$.

For $L=\left(\begin{array}{cc}-1/R& 1\\ 0& -1/(4R)\end{array}\right)$ with $R>1$, put $s=\sqrt{1-R^{-2}}$ and $q=(5-3s)/(5+3s)$. The optimal energy amplification of a linear system reaches its maximum atThe maximizing initial orientation is ${\rho }^{+}$ and the final orientation is ${\rho }^{-}$ from the orientation interval for transient energy growth. Taking $R\to \infty$ gives the displayed viscous-time and Reynolds-squared asymptotics. These constants belong to this specified triangular model, not to every non-normal matrix.

For $L=\left(\begin{array}{cc}-1/R& 1\\ 0& -1/(4R)\end{array}\right)$ and $R>1$, write $\rho =x_2/x_1$. The energy derivative is $x_1^2[-1/R+\rho -{\rho }^2/(4R)]$, positive exactly between the displayed orientations. Linear trajectories cross the sector from ${\rho }^{+}$ to ${\rho }^{-}$ because $\dot{\rho }=\rho [3/(4R)-\rho ]$. An energy-neutral rotational nonlinearity can reverse the crossing direction; the growth sector itself remains the same.

A nonlinear term $f(|x|)Nx$ with real skew-symmetric matrix $N$ contributes zero to the derivative of the quadratic energy $|x{|}^2/2$. It changes orientation rather than magnitude directly. If a separate linear generator is a non-normal matrix, that reorientation can still alter later transient growth, so energy cancellation alone does not make the nonlinear evolution equivalent to the linear one.

Every finite-dimensional complex normal matrix has an orthonormal eigenbasis, equivalentlyfor some unitary $U$ and diagonal $D$.