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Given a $2\times 2$ matrix $$ A=\begin{bmatrix} a&b\\ c&d\end{bmatrix} $$ Prove that$$\begin{Vmatrix} a&b\\ c&d\end{Vmatrix}=\sqrt{\frac{|A|^2+\sqrt{|A|^4-4(det(A))^2}}{2}}$$ |A| is defined as$\sqrt{a^2+b^2+c^2+d^2}$ I tried to let the input be $(\cos t,\sin t)$ ,but I can’t compute the maximum of $f(t)$

zx Z
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    what is |A| here ? – Lelouch Feb 23 '23 at 15:37
  • |A| is the length of A,it’s the definition of my textbook,It’s the square root of the sum of the square of the all elements in A,in this case |A|=The square root of(a^2+b^2+c^2+d^2) – zx Z Feb 23 '23 at 15:42

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The spectral norm of a matrix $A$ is given by $||A|| = \sqrt{\lambda_{\text{max}}(A^*A)}$, i.e. the square root of the largest eigenvalue of $A^*A$. See Wikipedia.

The eigenvalues of a $2\times2$ matrix can be expressed in terms of the trace and determinant: $\lambda_\pm = \frac{1}{2}\left(\textrm{tr} \pm \sqrt{\textrm{tr}^2-4\det}\right)$. This can be derived straightforwardly from solving the characteristic polynomial.

Thus, $$||A|| = \sqrt{\frac{1}{2}\left(\textrm{tr}(A^*A) + \sqrt{\textrm{tr}^2(A^*A)-4\det(A^*A)}\right)} \\[6pt] = \sqrt{\frac{|A|^2+\sqrt{|A|^4-4(\det(A))^2}}{2}} $$

Andreas Tsevas
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  • Maybe it's not necessary to use a general result for a $2\times2$ matrix. – Lelouch Feb 23 '23 at 15:54
  • Thank you for answering this question,but the norm in this problem is sup{|Av|},It’s not about the eigenvalue of a matrix ,or can you prove that the two definitions are equivalent? – zx Z Feb 23 '23 at 16:05
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    @zxZ You can find a proof of why the two definitions are equivalent in this post: Why does the spectral norm equal the largest singular value?. But it's true that using this general result for a $2\times 2$ matrix is a bit overkill. – Andreas Tsevas Feb 23 '23 at 16:09
  • @Andreas Now the problem is solved,thank you very much – zx Z Feb 23 '23 at 16:13
  • Is there any other solutions to compute ||A|| directly (using the definition given in this problem) i.e. ,computing the maximum of (ax+by)^2+(cx+dy)^2 for the input unit vector(x,y) – zx Z Feb 23 '23 at 16:17
  • @zxZ In principle, it should be possible to compute that maximum directly, e.g. by using Lagrange multipliers to enforce $x^2+y^2=1$. Else, your idea, with letting $(x,y)=(\cos t,\sin t)$ should also work. It will just involve very cumbersome calculations. – Andreas Tsevas Feb 23 '23 at 16:27