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$A\in M_3(\mathbb R)$ orthogonal matrix with $\det (A)=1$.

I need to prove that $(\mathrm{tr} A)^2-\mathrm{tr}(A^2) = 2 \mathrm{tr} (A)$ ; $\mathrm{tr}$=trace.

I know that if $A$ is orthogonal than $A^tA=I$ and that $A$ is diagonalizable and similar to $D=\begin{pmatrix} \lambda_1 & & \\ & \lambda_2 & \\ & & \lambda_3 \end{pmatrix}$. We know as well that $\mathrm{tr} A=\mathrm{tr} D= \lambda_1+ \lambda_2+\lambda_3$ that $\mathrm{tr} A^2=\mathrm{tr} D^2= \lambda_1^2+ \lambda_2^2+\lambda_3^2$ and that $\lambda_1 \lambda_2 \lambda_3=1$. It's not enough for solving the question.

What more should I know or use in order to solve it?

Thanks

Davide Giraudo
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Jozef
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5 Answers5

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$\begin{align*}\mathrm{Tr}(A)^2-\mathrm{Tr}(A^2)&= (\lambda_1+\lambda_2+\lambda_3)^2-(\lambda_1^2+\lambda_2^2+\lambda_3^2)\\&=2(\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1)\\&=2\lambda_1\lambda_2\lambda_3\left(\frac{1}{\lambda_1}+\frac{1}{\lambda_2}+\frac{1}{\lambda_3}\right)\\&=2\det(A)\mathrm{Tr}(A^{-1})\\&=2\mathrm{Tr}(A). \end{align*}$

(Edit: In the last step, we use the facts that $A^{-1}=A^T$ when $A$ is orthogonal and $\mathrm{Tr}(A^T)=\mathrm{Tr}(A)$ for any matrix $A$.)

anon
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  • Does this claim on $tr(A^{-1})$ is always true? – Jozef Aug 23 '11 at 09:38
  • @anon: You could add that your last step holds because $A^{-1}=A^t$ and $\text{tr}(A^t)=\text{tr}(A)$. – Did Aug 23 '11 at 09:42
  • Yeah, I know this I mean that for every $A$ such that $trA= x+y+z+w$, $trA^{-1}=1/x+1/y+1/z+1/w$? – Jozef Aug 23 '11 at 09:43
  • Jozef: The eigenvalues of $A^n$ are the $n$-th powers of the eigenvalues of $A$, so $\mathrm{Tr}(A^n)=\sum \lambda_i^n$ holds. – anon Aug 23 '11 at 09:47
  • +1: This is the most elegant use of the orthogonality of $A$ (out of the solutions presented here at this point in time). Nice! – Jyrki Lahtonen Aug 23 '11 at 10:47
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In $\mathbb Z[X,X^{-1}]$ we have $$(1+X+X^{-1})^2-(1+X^2+X^{-2})=2+2X+2X^{-1}.$$ In other words, your equality holds over any commutative ring, for any matrix similar to a diagonal matrix with diagonal entries $1,a,a^{-1}$, where $a$ is a unit.

Pierre-Yves Gaillard
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An old-fashioned argument

The matrix $A$ is the matrix of a rotation through some angle $\theta$. It is a standard result that the trace of such a rotation is $1+2\cos\theta$. The matrix $A^2$ is a rotation through $2\theta$ about the same axis.

So we want to show that $$(1+2\cos\theta)^2-(1+2\cos 2\theta)=2(1+2\cos\theta).$$

With a little manipulation this reduces to the familiar $\cos 2\theta=2\cos^2\theta -1$.

André Nicolas
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A rotation is diagonalisable over the complex numbers, but generally not over the reals. However, there always is an orthonormal basis $\mathcal{B}$ and a real number $\theta$ such that $$\mathrm{Mat}_{\mathcal{B}}(A)=\left(\begin{array}{ccc} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{array}\right)$$ and you'll have $$\mathrm{Mat}_{\mathcal{B}}(A^2)=\left(\begin{array}{ccc} \cos(2\theta) & -\sin(2\theta) & 0 \\ \sin(2\theta) & \cos(2\theta) & 0 \\ 0&0&1\end{array}\right)$$ Thus $\mathrm{Tr}(A)=1+2\cos(\theta)$ and $\mathrm{Tr}(A^2)=1+2\cos(2\theta)$. The rest follows from trigonometry.

Olivier Bégassat
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Hint: $$\det(I-A)=\det(A^tA-A)=\det(A^t-I)\det(A)=\det(A-I)\cdot1=(-1)^3\det(I-A),$$ so $\det(I-A)=0$. IOW one of the eigenvalues is equal to $1$. Does that help?

Jyrki Lahtonen
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  • Hi Jyrki! Isn't it enough to note that the eigenvalues are $1,z,1/z$ with $|z|=1$? (It's a Laurent polynomial identity, i.e. an identity in the ring $\mathbb Z[z,z^{-1}]$.) – Pierre-Yves Gaillard Aug 23 '11 at 09:21
  • @Pierre-Yves: Of course it is enough! Undoubtedly you know this, but for the record I will state that geometrically $A$ is a rotation of the 3-dimensional space. Hence (by the above argument) it has an axis (corresponding to eigenvalue 1), and an angle of rotation $\theta$ (meaning that the other two eigenvalues are $\lambda_{1,2}=\cos\theta\pm i\sin\theta.$) – Jyrki Lahtonen Aug 23 '11 at 09:27