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Let us consider the euclidean norm on $\mathbf{R}^2$. After some computations, I have obtained the following expression for the associated operator norm on 2 by 2 matrices.

$$ \left\lVert\pmatrix{a&b\cr c&d\cr}\right\rVert^2 = {1\over 2} \Bigl(\lvert a+ib\rvert^2+\lvert c+id\rvert^2+\lvert(a+ib)^2+(c+id)^2\rvert\Bigr). $$

This expression is new to me and I am wondering if there is a conceptual explanation for such a formula. Also, is there an analogous formula for higher dimensional matrices?

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  • $\begingroup$ You specify a norm on $\mathbf R^2$, but not on $2\times2$ matrices. Do you mean to equip matrices with the operator norm? $\endgroup$
    – LSpice
    Commented Feb 9, 2022 at 23:41
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    $\begingroup$ @LSpice - I believe that "induced norm" implies "operator norm". $\endgroup$
    – Nathaniel Johnston
    Commented Feb 9, 2022 at 23:45
  • $\begingroup$ Yes, this is the operator norm associated to the euclidean norm. Edited. $\endgroup$
    – coudy
    Commented Feb 10, 2022 at 10:03
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    $\begingroup$ For what it's worth: noting that $\|A\|^2$ is the largest eigenvalue of $A^{\mathrm{t}}\,A$, so it is a root of the characteristic polynomial of the latter should give the above formula, implies that for $n\times n$ matrices with rational entries, $\|A\|^2$ is algebraic of degree $n$ at most, and suggests that we probably can't do much better in general than maybe a factor $\frac{1}{2}$ over this. $\endgroup$
    – Gro-Tsen
    Commented Feb 10, 2022 at 10:46
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    $\begingroup$ Here is a specific example: $A = \begin{pmatrix}0&0&1\\0&1&1\\1&1&0\\\end{pmatrix}$ has $\|A\|^2 ≈ 3.25$ which is root of the irreducible polynomial $x^3 - 5x^2 + 6x - 1$ over $\mathbb{Q}$, so, algebraic of degree $3$. This suggests that it will be difficult to find an analogous formula for $3\times 3$ matrices. $\endgroup$
    – Gro-Tsen
    Commented Feb 10, 2022 at 10:56

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In essence, the answer is: no.

  1. $\|A\|^2$ is always equal to the largest eigenvalue of $A^tA$. If $A$ is already symmetric, $\|A\|$ is equal to the (absolute value of) the largest eigenvalue of $A$. The characteristic polynomial of a matrix can be essentially anything, so the matrix norm can be essentially any algebraic number. This means a nice expression with only basic arithmetic is impossible, and any somewhat nice expression is unlikely. Otherwise, linear algebra software would not go throught all the trouble of using complicated iterative numerical algorithms to approximate eigenvalues.

  2. But it is possible to extend (slightly) beyond $2\times2$. For example for a symmetric $3\times 3$ or $4\times 4$ matrix, you can compute the characteristic polynomial, and plug its coefficients into the formulas for the roots of these polynomials (i.e. Cardano or Ferrari), and then pick out the largest of the solutions. Note that this is actually done in practice, as it can be faster than the iterative methods that are otherwise used to compute eigenvalues of (large) matrices. See for example here for the special case of $3\times3$ matrices. Though already in this case, the formulas are rather cumbersome.

  3. If you are looking for bounds, the simplest ones are \begin{align} \frac{1}{\sqrt{n}}\|A\|_F \le \|A\| \le \|A\|_F \end{align} where \begin{align} \|A\|_F^2 = \sum_{i,j=1}^n a_{ij}^2 \end{align} is the "Frobenius norm" of the matrix. More precise bounds can be obtained using the Gershgorin circle theorem for example

  4. As a side note: because of all these problems, in numerical linear algebra, one rarely uses the $\ell_2$-induced matrix norm in actual calculations. For example the matrix-norms induces by the $\ell_1$ and $\ell_\infty$ norms have very simple expressions, so they are preferable in practical calculations.

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