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Is it true that $$\|A\| \leq \|A^2\|$$ for $A \in SL(2,\mathbb{R})$, where $\| \|$ is the operator norm that is the first singular value? $$\left \| A \right \| =\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A).$$

Definitively, it is not true for $GL(2, \mathbb{R})$ as one can consider $A=diag(1/3, 1/3).$

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    $\begingroup$ What's the Euclidean norm? Is that the Frobenius norm, i.e. the square root of the sum of squares of the entries? $\endgroup$ Oct 20, 2021 at 22:47
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    $\begingroup$ @TheoBendit : The first singular value. $\endgroup$
    – Adam
    Oct 20, 2021 at 22:49
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    $\begingroup$ The operator norm can be bigger than the largest absolute value of eigenvalues if the matrix is not diagonalizable, e.g. $\pmatrix{1&1\\0&1}$. $\endgroup$
    – Berci
    Oct 20, 2021 at 23:00
  • $\begingroup$ @Berci, maybe "singular value" is not "eigenvalue"? Perhaps the usage here is ambiguous... $\endgroup$ Oct 20, 2021 at 23:02
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    $\begingroup$ @JeppeStigNielsen : yes, it is, but how one can conclude whether my question is true or not? $\endgroup$
    – Adam
    Oct 20, 2021 at 23:35

2 Answers 2

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A counterexample could be $$A=\pmatrix{3&5\\-2&-3}.$$


Some thought about how you can see that this $A$ works.

First of all, its determinant is $-9-(-10)=1$ as required. This also means that the product of its two eigenvalues is $1$.

The trace of $A$ is $3-3=0$. So the sum of the eigenvalues is $0$. Given the product and the sum of the eigenvalues, it is clear that the eigenvalues are $+i$ and $-i$ where $i$ is the imaginary unit.

Since the eigenvalues are distinct, this $A$ is diagonalizable by some complex matrix in $GL(2,\mathbb{C})$.

Now think about $A^2$. It is diagonalizable as well (because $A$ is, and by the same complex matrix). The eigenvalues of $A^2$ are $-1$ and $-1$, the squares of the eigenvalues of $A$. But a diagonalizable matrix all of whose eigenvalues are equal, is really a diagonal matrix, so we have $$A^2=\pmatrix{-1&0\\0&-1}.$$

The operator norm you are asking about satisfies $$\|A\|=\sup\left\{\|Ax\|\,\middle|\, x\in\mathbb{R}^2 \text{ and } \|x\|=1\right\}$$ where the symbols $\|\cdot\|$ inside the set on the right-hand side denotes the standard (Euclidean) length of a vector in $\mathbb{R}^2$. So $\|A\|$ is the maximal length of the image of a unit vector.

It is clear that $\|A^2\|=1$ since $A^2$ maps all unit vectors to unit vectors. It is also clear that $\|A\|$ is strictly larger. For simplicity, take $x=\pmatrix{0\\1}$, then $Ax=\pmatrix{5\\-3}$ whose length is $\sqrt{5^2+3^2}=\sqrt{34}$, so it follows that the operator norm of $A$ is at least this number, $\|A\|\ge\sqrt{34}$.


These arguments show that any matrix $$A=\pmatrix{t&s\\-\frac{1+t^2}{s}&-t}$$ where $s\ne 0$ and $t$ are real numbers has determinant $1$ and trace $0$ and therefore will square to $\pmatrix{-1&0\\0&-1}$. By taking both $s$ and $t$ huge, you can make the length of the second column of $A$ as huge as you want. So there are matrices with arbitrarily large operator norms in $SL(2,\mathbb{R})$ whose squares have operator norm $1$.

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    $\begingroup$ Thank you very much for the example and explanation. I got one more question: do you think the statement is true for hyperbolic matrices? $A \in SL(2,\mathbb{R})$ is hyperbolic if $|\text{trace}(A)| >2.$ $\endgroup$
    – Adam
    Oct 21, 2021 at 15:34
  • $\begingroup$ @Adam I do not know now. I wonder if it could be true for $A\in SL(2,\mathbb{R})$ if only $\text{trace}(A)\ne 0$. $\endgroup$ Oct 21, 2021 at 19:17
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    $\begingroup$ No! that is silly. You can make a small "perturbation"/adjustment of one diagonal entry in my matrix $A$ above and have an example where the trace is nonzero. But I still think you can be right with $|\text{trace}(A)|>2$. $\endgroup$ Oct 21, 2021 at 19:58
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If by "singular values" you mean the diagonal entries in writing $A=k_1 \pmatrix{t&0 \cr 0&t^{-1}}k_2$, with $k_1, k_2$ in the orthogonal group $SO(2)$, then your speculation is correct. EDIT: if you truly mean "spectral norm", then this is not correct. But/and in many contexts "singular values" does not refer to eigenvalues. Dunno... Let me continue under the assumption that you do not actually mean eigenvalues or spectral norm, but as I wrote... (someone may be interested in that scenario, in any case).

EDIT: this asserts nothing about diagonalizability of $A$. The general version is "Cartan decomposition". Then, letting $\delta=\pmatrix{a&0\cr 0&d}$, $$ A^\top A \;=\; (k_1 \delta k_2)^\top(k_1Ak_2) \;=\; k_2^\top \delta^\top k_1^\top k_1 \delta k_2 \;=\; k_2^\top \delta^2 k_2 $$ There are certainly variations depending on whether one wants to consider complex matrices, unitary versus orthogonal, and so on.

From such an expression, visibly, the singular values (EDIT: if this is truly what is intended) of the square are the squares of the singular values. For $SL_2(\mathbb R)$, the two singular values must be $t$ and $t^{-1}$. Whichever is larger, when squared, is the larger of the two squares. Since the maximum of $t$ and $t^{-1}$ is always at least $1$, the claim follows.

(Thanks to people trying to improve my answer, etc.)

EDIT-EDIT: (as @Jeppe Stig Nielsen notes in a comment), indeed, the operator norm is the square root of the largest eigenvalue of $A^\top A$, etc. Which is not so much related to the eigenvalues of $A$ itself, but to its "singular values" in an Iwasawa decomposition (see above). Perhaps this observation will resolve some confusion. (Believe me, these seemingly-slightly-different versions of similar things confounded me for a long time... :)

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    $\begingroup$ Thanks for your answer. I don't understand why you wrote $A=k_1 diag(a, d) k_2$, the definition is $$\left \| A \right \| =\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A)$$. One more thing: what do you mean by $t$ and $t^{-1}?$ $\endgroup$
    – Adam
    Oct 20, 2021 at 23:11
  • $\begingroup$ you diagonalized $A$? if so, it shouldn't be $A=k_1 diag(a, d) k_1^{-1}$? $\endgroup$
    – Adam
    Oct 20, 2021 at 23:20
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    $\begingroup$ There is no reason why an $A\in SL_2(\mathbb{R})$ should be diagonalizable (let alone orthogonally diagonalizable). $\endgroup$ Oct 20, 2021 at 23:40
  • $\begingroup$ @Adam, if your comment truly gives the definition of the norm that you want, it has some self-contradiction. That is, that square-root expression is not reliably the (absolute value of) largest eigenvalue of $A$. That square-root expression is what my response addresses. "Diagonalizing" would indeed mean what your comment suggests, but that's not what's involved in the thing expressed by your square root expression (nor my response). And, again, the interpretation I'm taking assumes nothing about diagonalizability of $A$, nor about eigenvalues, etc. $\endgroup$ Oct 21, 2021 at 0:51
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    $\begingroup$ As the question stands now, I guess it is clear enough that $\| A \|$ was meant to mean the operator norm of $A$ (with respect to the standard Euclidean norm of vectors in $\mathbb{R}^2$). And in that case it is true that $\| A \|$ is also the square root of the largest eigenvalue of $A^T A$. $\endgroup$ Oct 21, 2021 at 0:57

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