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Let $L: V \rightarrow V$ be a linear operator on a finite dimensional real inner product space $V$ such that $L^{*} = L^{3}$. Show that $L^{2}$ is diagonalizable over $\mathbb{R}$.

Attempt: Suppose $L^{*} = L^{3}$. Then $LL^{*} = LL^{3} = L^{3}L = L^{*}L$. It follows that $L$ is a normal. By the real spectral theorem (I am only used to the complex version), there exists an orthonormal basis of $V$ consisting of eigenvectors of $L$, say $\{v_{1},..., v_{n} \}$. But then this consists of eigenvectors of $L^{2}$ since $Lv_{j} = \lambda_{j}v_{j} \implies L^{2}v_{j} = \lambda_{j}^{2}v_{j}$ and so $L^{2}$ is diagonalizable.

Is my use of the spectral theorem legal?

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  • $\begingroup$ $L=PDP^{-1}$ implies $L^2 = PD^2 P^{-1}$. $\endgroup$ – i707107 Dec 11 '16 at 17:30
  • $\begingroup$ Yes, this is correct and it is essentially my conclusion. I thought the fact that the ground field is $\mathbb{R}$ rather than complex would require a different argument. All I did was to simply invoke the real case of the spectral theorem. $\endgroup$ – akech Dec 11 '16 at 17:33
  • $\begingroup$ I am also not sure about if we shall apply spectral theorem for complex case directly. So yet another approach is to show that $L^2 = (L*)^2$, since for real inner product space, self adjoint means othrogonally diagonalizable…but I am trying so see if $L^2$ is self adjoint $\endgroup$ – Li Chun Min Dec 11 '16 at 18:15
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    $\begingroup$ @LiChunMin any normal operator with real eigenvalues is self-adjoint $\endgroup$ – Omnomnomnom Dec 11 '16 at 18:29
  • $\begingroup$ @Omnomnomnom oh yes those are real haha $\endgroup$ – Li Chun Min Dec 11 '16 at 18:35
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You have the right basic idea, but you have yet to prove the conclusion. We can extend the real vector space $V$ into a complex inner product space $\tilde V$. $L$ is unitarily diagonalizable over $\tilde V$ by the spectral theorem, which means that $L^2$ is also unitarily diagonalizable over $\tilde V$ (since it has the same eigenvectors).

In order to conclude that $L^2$ is diagonalizable over the real vector space $V$, we must first conclude that its eigenvalues are real. In order to do so, note that since $L^* = L^3$ and $L$ is (untiarily!!) diagonalizable, each eigenvalue $\lambda$ of $L$ satisfies $\overline{\lambda} = \lambda^3$, and therefore conclude that the only possible eigenvalues of $L$ are...?

Alternatively: note that $(L^2)^2 = L^*L$ necessarily has non-negative eigenvalues, from which we may conclude that $L^2$ has real eigenvalues.

From there, it suffices to note that for a real $\mu$, $\ker (L^2 - \mu I)$ has the same dimension as a real subspace of $V$ as it does as a complex subspace of $\tilde V$.

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I am not sure the use of the spectral theorem for L is correct. For $L$ normal it states that: $L=ULU^*$ where $U$ is unitary, i.e. complex, this cannot be restricted to $\mathbb{R}$.

An example of such a linear operator would be: $[{\begin{array}{cc} 2 & -3 \\ 3 & 2 \ \end{array} }] $. It is a real normal matrix but cannot be diagonalized into real values.

Instead, the approach I see is to realize $M=LL^*=L^4$ is symmetric. Applyting the spectral theorem yields it can be diagonalized into real eigenvectors and eigenvalues. Besides, being an even power the matrix will be positive semidefinite. This means that:

$ LL^* = L^4 = SDS^{-1} $

$L^2 = S \sqrt{D} S^{-1} $

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  • $\begingroup$ There's nothing which guarantees that $L^2$ is positive deifinite, even if it is a square root of $L^4$. $\endgroup$ – Omnomnomnom Dec 11 '16 at 18:35
  • $\begingroup$ $L$ is normal so it is diagonalizable. It can be expressed as $L=S D_L S^{-1}$ where $D_L$ is diagonal matrix different from D. Then $L^2=S D_L^2 S^{-1}$ then $D_L^2=\sqrt{D}$ and since it is the square of $D_L$ it has to be positive semidefinite. The approach of the symmetric matrix is used to show that eigenvalues and eigenvectors are real. $\endgroup$ – ebabio Dec 11 '16 at 20:43
  • $\begingroup$ Note that $$ \pmatrix{1&1\\0&-1}^2 = I $$ but $$ \pmatrix{1&1\\0&-1} \neq \sqrt{I} $$ what's worse, the matrix on the left isn't even normal $\endgroup$ – Omnomnomnom Dec 11 '16 at 20:49
  • $\begingroup$ Or, consider the case of $$ L^2 = \pmatrix{1&0\\0&-1} $$ $\endgroup$ – Omnomnomnom Dec 11 '16 at 20:59
  • $\begingroup$ Or I guess it depends what you mean by $\sqrt{}$. Usually, it refers specifically to the positive (or positive definite) square root. $\endgroup$ – Omnomnomnom Dec 11 '16 at 21:00

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