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Given an $n \times n$ normal matrix $A$ over $\mathbb{R}$, show that there is an orthogonal matrix $O$ such that $O^TAO$ is either diagonal or in the form

$$\left[\begin{matrix}{\lambda}&0&0\\0&a&b\\0&-b&a\end{matrix}\right].$$

So far, my thoughts are as follows. The characteristic polynomial $f$ of $A$ is cubic, so it has at least one real root, call it $\lambda$ with corresponding eigenvector $V_\lambda$, and $f = (x-\lambda)q$ for some polynomial $q$ of degree 2.

If $q$ has distinct real roots, then $A$ has an orthonormal basis of eigenvectors, which give the columns of a matrix $O$ such that $O^{-1}AO = O^TAO$ is diagonal.

Otherwise, ...? Essentially all of the theorems and exercises in my text work over $\mathbb{C}$, not $\mathbb{R}$. If we were working over $\mathbb{C}$, we would always have an orthonormal basis of eigenvectors $A$ would always be diagonalizable by such a matrix $O$.

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You need to work out the details that the $2$-dimensional invariant subspace corresponding to eigenvalues $\lambda,\bar\lambda$ for a complex (non-real) eigenvalue $\lambda$, with the appropriate basis, gives the matrix $\begin{bmatrix} a&b\\-b&a\end{bmatrix}$ when $\lambda = a+ib$. (Hint: If $v\in\Bbb C^3$ is an eigenvector with eigenvalue $\lambda$ of the complexification of $A$, write $v=x+iy$ for $x,y\in\Bbb R^3$.)

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  • $\begingroup$ Following this hint, $A$ has an orthonormal basis of eigenvectors over $\mathbb{C}^3$ with one real eigenvalue $\lambda_R$ and corresponding eigenvector $v_R$. Let $\lambda=a+bi$, $v = x+yi$; $Av = \lambda v \implies Au=au-bv$, $Av=av+bu$. Taking ordered basis $\mathcal{B}=\{v_R,u,v\}$ and using it as colunns of $O$, we find that the matrix of $A$ in$\mathcal{B}$ is $\left[\begin{matrix}{\lambda}&0&0\\0&a&b\\0&-b&a\end{matrix}\right]$. Is this correct? $\endgroup$ – user169845 Jan 4 '15 at 22:10
  • $\begingroup$ Looks good. You should stop to think about how the unitary basis for $\Bbb C^3$ translates to a basis for $\Bbb R^3$. If you follow this, do you have a (real) orthonormal basis? $\endgroup$ – Ted Shifrin Jan 4 '15 at 22:26
  • $\begingroup$ @user169845: Remember that you need to check that $u,v$ are linearly independent over the reals. $\endgroup$ – copper.hat Jan 4 '15 at 23:49

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