$\DeclareMathOperator\SL{SL}\DeclareMathOperator\trace{trace}$Let $A \in \SL(2,\mathbb{R})$ and $\trace(A)>2$. Is it true that $$\lVert A\rVert \leq \lVert A^2\rVert,$$ where $\lVert \rVert$ is the operator norm that is the first singular value? $$\lVert A \rVert =\sqrt{\lambda_{\text{max}}(A^*A)}=\sigma_{\text{max}}(A).$$

Let me mention that if the condition $\trace(A)>2$ is removed, then the above statement is not true; see Jeppe Stig Nielsen's answer to Is it true that $\lVert A\rVert \leq \lVert A^2\rVert$ for $A\in \operatorname{SL}(2, \mathbb{R})$? on MSE.

My attempt: I think it is true: The operator norm satisfied $$\lVert A\rVert=\sup\left\{\lVert Ax\rVert \,\middle\vert\, \text{$x\in\mathbb{R}^2$ and $\lVert x\rVert=1$}\right\}$$ where the symbols $\lVert \rVert$ inside the brackets on the right-hand side denote the standard (Euclidean) length of a vector in $\mathbb{R}^2$. So $\lVert A\rVert$ is the maximal length of the image of a unit vector. On the other hand, $\trace(A)>2$, so one of the eigenvalues is greater than the other one.

  • $\begingroup$ Your link 'here' was to edit your own post. I figured you probably wanted to link to the answer with a counterexample. Also, you use $A^{^*}$ A^{^*} for the adjoint; it should be $A^*$ A^*. I edited accordingly. $\endgroup$
    – LSpice
    Commented Oct 24, 2021 at 17:02
  • 1
    $\begingroup$ It is considered inappropriate to completely edit your question after it has been answered. Instead, if you have another question, ask it separately. $\endgroup$ Commented Oct 25, 2021 at 21:50
  • $\begingroup$ @SamHopkins : I want to delete it as I made a big mistake to ask the question; I am really regretful. Could you please delete it? $\endgroup$
    – Adam
    Commented Oct 25, 2021 at 22:02
  • 1
    $\begingroup$ @Adam: again, it is considered inappropriate to delete a MO question after it has been answered (and anyways if you try you'll see you can't). In unusual/extreme cases you should get in contact with a moderator (which I am not, but Stefan Kohl mathoverflow.net/users/28104/stefan-kohl - who was also reverting your edits - is). $\endgroup$ Commented Oct 25, 2021 at 22:05

1 Answer 1


We can do this by a calculation. The assumptions on the determinant and trace are equivalent to having eigenvalues $\lambda,1/\lambda$, with $\lambda>1$. We can rotate the first eigenvector to the $e_1$ position, and then $$ A=\begin{pmatrix} \lambda & b \\ 0 & 1/\lambda \end{pmatrix} , $$ so $$ A^*A=\begin{pmatrix} \lambda^2 & \lambda b \\ \lambda b & b^2 +1/\lambda^2 \end{pmatrix} $$ In general, the eigenvalues of a $B\in\textrm{SL}(2,\mathbb R)$ are $T/2 \pm \sqrt{T^2/4-1}$, with $T=\textrm{tr}\; B$.

In the case of $A^*A$, we are interested in the large eigenvalue (obtained for $+$). Clearly, this is increasing in $T$. So now the question is if $(A^2)^*A^2$ has a larger trace $T_2$, and the same calculation now gives $$ T_2 = \lambda^4 + \frac{1}{\lambda^4} + b^2 \left( \lambda+\frac{1}{\lambda}\right)^2 . $$ This indeed satisfies $T_2\ge \lambda^2+1/\lambda^2 + b$. In fact, we have $\lambda^4+1/\lambda^4\ge \lambda^2+1/\lambda^2$ for $\lambda\ge 1$, or, equivalently, $f(x)\equiv x^4-x^3-x+1\ge 0$ for $x\ge 1$, since $f'(x)\ge 0$ in this range and $f(1)=0$.

  • $\begingroup$ Thank you very much for your answer. Could you please clarify the first paragraph of your answer ? Thanks in advance. $\endgroup$
    – Adam
    Commented Oct 24, 2021 at 19:20
  • $\begingroup$ As far as I know, the matrix $A$ is conjugated to the diagonal matrix $D=diag(\lambda, 1/ \lambda)$, which means $A=PDP^{-1}$. I don't understand why you wrote $ A=\begin{pmatrix} \lambda & b \\ 0 & 1/\lambda \end{pmatrix} . $. $\endgroup$
    – Adam
    Commented Oct 24, 2021 at 19:34
  • $\begingroup$ @Adam: This is the general matrix with $Ae_1=\lambda e_1$ and $\det A=1$. This form is more useful than $A=PDP^{-1}$ because it only has one additional parameter ($b$) rather than a full matrix ($P$). $\endgroup$ Commented Oct 24, 2021 at 19:38
  • $\begingroup$ @ Christian Remling That is interesting; I haven't seen it before. Could you please send a reference where I can see its proper proof? Thanks in advance $\endgroup$
    – Adam
    Commented Oct 24, 2021 at 21:22
  • 1
    $\begingroup$ @Adam: The matrix representation of a linear map is with respect to a basis. Now you can just say that you choose as your first basis vector the eigenvector of the large eigenvalue (and take the orthogonal direction as the second basis vector, so that you have an ONB and can still compute operator norms in the usual way). Alternatively (essentially the same argument really), replace $A$ by $RAR^{-1}$, with $R$ chosen as a rotation matrix that moves that eigenvector to $e_1$. Since $R$ is orthogonal, conjugation by $R$ does not change the operator norm. $\endgroup$ Commented Oct 25, 2021 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.