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I have a linear operator $T$ which acts on the vector space of square $N\times N$ matrices in this way:

$T(A)=0.5(A-A^\mathrm{t})$

($A^\mathrm{t}$: the transpose of a matrix $A$).

I need to prove that this operator is diagonalizable.

I know that a linear transformation is diagonalizable if and only if there is a basis for the vector space that consists entirely of eigenvectors.

I want to find the possible eigenvalues for this operator, so by definition I need to see what are the $\lambda$s that imply $t(v)=\lambda v$,and by assignment I get: $A(1-2\lambda)=A^\mathrm{t}$.

How can I go on from here?

Thank you guys.

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  • $\begingroup$ All square matrices? Or the square matrices of a given size? $\endgroup$ – Arturo Magidin Mar 17 '11 at 15:01
  • $\begingroup$ All square matrices NxN. $\endgroup$ – user6163 Mar 17 '11 at 15:02
  • $\begingroup$ Avnon: So it's "all square matrices of a given size." $\endgroup$ – Arturo Magidin Mar 17 '11 at 15:08
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An alternative way to proceed if you don't know about symmetric and skew-symmetric matrices (or projections) and you don't want to try to find the characteristic polynomial for arbitrary $N$ is the following:

Start with the general observation: if $\mathbf{x}$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda$, then $\mathbf{x}$ is also an eigenvector of $T^2$ ($T$ composed with itself) corresponding to $\lambda^2$:

Indeed, we have that $\mathbf{x}\neq\mathbf{0}$ (since $\mathbf{x}$ is an eigenvector of $T$), and \begin{align*} T^2(\mathbf{x}) &= T(T(\mathbf{x})) = T(\lambda\mathbf{x})\\ &= \lambda T(\mathbf{x}) = \lambda(\lambda\mathbf{x}) = \lambda^2\mathbf{x}. \end{align*}

Caveat. The converse does not hold: you can have $\mathbf{x}$ be an eigenvector of $T^2$, but not of $T$: consider the rotation of the real plane by ninety degrees; then ever vector is an eigenvalue of $T^2$, but $T$ has no eigenvectors.

So, consider your $T$. If you apply $T$ twice, you get: $$T(T(A)) = T\left(\frac{1}{2}(A-A^t)\right) = \frac{1}{2}\left(\frac{1}{2}(A-A^t) - \frac{1}{2}(A^t-A)\right) = \frac{1}{2}(A-A^t) = T(A).$$ So if $A$ is an eigenvector if corresponding eigenvalue $\lambda$, then you have that $T(A)=\lambda A$ and $T^2(A)=\lambda^2A$. But since $T(A)=T^2(A)$, then $\lambda A = \lambda^2 A$, hence $(\lambda-\lambda^2)A=\mathbf{0}$. Since $A\neq\mathbf{0}$, then $\lambda=\lambda^2$, so either $\lambda=0$ or $\lambda=1$.

So the only possible eigenvalues of $T$ are $\lambda=0$ and $\lambda=1$. (Note: This is true of any linear transformation $T$ such that $T=T^2$; these are called "projections").

Now that we know the possible eigenvalues, it's a bit easier to check. If $A$ is an eigenvector corresponding to $0$, then we must have $ 0 = \frac{1}{2}(A-A^t)$, so $A=A^t$; that is, $A$ must be equal to its transpose (symmetric)

If $\lambda=1$, then we must have $A = \frac{1}{2}(A-A^t)$, so $A=-A^t$; that is, $A$ must be skew-symmetric.

So this determines all the eigenvectors and eigenvalues of $T$. It is now straightforward to verify that you can find enough linearly independent symmetric and skew-symmetric matrices (to wit, $n^2$ of them) to get a basis for the vector space, so that $T$ is diagonalizable.

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  • $\begingroup$ Thank you so much. Very helpful answer. $\endgroup$ – user6163 Mar 22 '11 at 10:25
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$T$ is the projection onto to the space of antisymmetric matrices along the symmetric matrices. Thus the spectrum is $\{0;1\}$, and those are the eigenspaces. Obviously they span the whole space, so $T$ is diagonalizable.

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  • $\begingroup$ hey, thanks for the comment. what do you mean by the spectrum is {0;1}? I'll be glad for a little more explanation. thanks. $\endgroup$ – user6163 Mar 17 '11 at 15:47
  • $\begingroup$ @Nir: "Spectrum" means the set of eigenvalues. In other words, the eigenvalues are $\lambda=0$ (for $A$ symmetric) and $\lambda=1$ (for $A$ antisymmetric). $\endgroup$ – Hans Lundmark Mar 17 '11 at 15:51
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This is a funny problem because although $T(A)$ is the skew symmetric part of $A$, $T$ itself has a matrix representation that is symmetric.

In particular, let $E_{nm}= \left (\delta_{nm} \right)$, i.e. the matrix of 0s everywhere except a 1 at the $(n,m)$ entry. Then $T(E_{nm}) =\frac{1}{2} \left(E_{nm}-E_{mn}\right)$. With the usual inner product on $N\times N$ matrices, we have that $$T=\frac{1}{2} \sum_{n,m=1}^{N} (E_{nm}\otimes E_{nm} -E_{mn}\otimes E_{nm}).$$ So $$T^{\top} = \frac{1}{2} \sum_{n,m=1}^{N} (E_{nm}\otimes E_{nm}-E_{nm}\otimes E_{mn})=T$$ (by invariance under change of notation ;).

Added: Since real symmetric matrices are diagonalizable, $T$ is.

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