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Let $A=\begin{pmatrix} 1& 1 \\ a^2 &1 \end{pmatrix} \text{ with } a\in (0,\frac{1}{2}]$. Show $$cond_2(A)=||A||_2 \cdot ||A^{-1}||_2\leq 4(1-a^2)^{-1}$$ by first showing $||A||^2_2\leq||A||_1||A||_{\infty}$.

$||A||^2_2$ is the maximal eigenvalue of $A^TA$ and $||A||_1||A||_{\infty}=2\cdot2 = 4$. Prove $||A||^2_2\leq||A||_1||A||_{\infty}=4$ by contradiction: Suppose $\lambda_{max} >4$, then $$4<\lambda_{\ast}=\frac{\sqrt{a^8+2a^4+4a^2+1}+a^4+3}{2} \iff 5<\sqrt{a^8+\underbrace{2a^4}_{\leq \frac{1}{2}}+\underbrace{4a^2}_{\leq 1}+1}+\underbrace{a^4}_{\leq 1} \\ \leq \sqrt{4}+1 = 3$$ contradiction!

So I know $||A||^2_2\leq 4 \Rightarrow ||A||_2\leq 2$ but I need to show $cond_2(A) = \underbrace{||A||_2}_{\leq 2}||A^{-1}||_2\leq 4(1-a^2)^{-1}$. I don't know why I needed to show $||A||^2_2\leq||A||_1||A||_{\infty}$ first? How does it help? I still need the maximal eigenvalue of $(A^{-1})^TA^{-1}$. The eigenvalues are $$\lambda_1=\frac{\sqrt{a^4-2a^2+5}-a^2+1}{2\sqrt{a^4-2a^2+5}-4} \\ \lambda_2=\frac{\sqrt{a^4-2a^2+5}-a^2-1}{2\sqrt{a^4-2a^2+5}+4}$$ What am I supposed to do now? Thanks for any help!

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2 Answers 2

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If $\lambda_1,..., \lambda_n$ are the eigenvalues of an invertible $n\times n$ matrix $M$, then $\frac{1}{\lambda_1},\dots,\frac{1}{\lambda_n}$ are the eigenvalues of $M^{-1}$. Moreover, the eigenvalues of $M^T$ are the same as the eigenvalues of $M$.

Putting this together, we have that the eigenvalues of $(A^{-1})^{T} A^{-1}$ are the reciprocals of the eigenvalues of $((A^{-1})^{T} A^{-1})^{-1} = AA^T$, which in turn are the same as the eigenvalues of $(AA^T)^T = A^T A$. Moreover, since the eigenvalues of $A^T A$ are necessarily non-negative, we get that $\|A^{-1}\|_2^2 = \frac1{\lambda_{\min}(A^T A)}$. Now, the determinant of a matrix is just the product of its eigenvalues, so from a direct computation:

$$(1 - a^2)^2 = \det(A^T A) = \lambda_{\max}(A^T A) \cdot\lambda_{\min}(A^T A) = \frac{\|A\|_2^2}{\|A^{-1}\|_2^2}.$$ Hence $\|A^{-1}\|_2 = \frac{\|A\|_2}{1 - a^2}$, and so $\text{cond}_2(A) = \frac{\|A\|_2^2}{1 - a^2} \le \frac{4}{1 - a^2}$.

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  • $\begingroup$ Thank you! I determined $\lambda_{min} (A^TA)=\frac{a^4+3-\sqrt{a^8+2a^4+4a^2+1}}{2}$. So it is $cond_2(A)=||A||_2||A^{-1}||_2\leq 2 \cdot \frac{2}{a^4+3-\sqrt{a^8+2a^4+4a^2+1}}$ and I need to show $(1-a^2)\leq a^4+3-\sqrt{a^8+2a^4+4a^2+1} \iff \sqrt{a^8+2a^4+4a^2+1}\leq \sqrt{4}<2+a(a^2+1)$ since $a>0$. So the inequality should be proved, shouldn't it? $\endgroup$ May 19 at 20:29
  • $\begingroup$ @PhilSpa I made the mistake of putting $\|A\|_2 = \lambda_{\max}(A^T A)$ instead of $\|A\|_2^2$. With that correction, the trace argument falls apart. However, we can argue very similarly using the determinant, and with the corrected norm the answer follows immediately. I've edited the answer to include that argument. $\endgroup$ May 20 at 10:05
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$0 \lt 1-a^2 = \det\big(A)=\lambda_1\cdot \lambda_2 = \sigma_1\cdot \sigma_2$
i.e. the product of singular values gives the modulus of the determinant (and in particular when the determinant positive it is equal to the product of singular values)

proof: using polar decomposition
$\det\big(A\big) =\det\big(UP\big)=\det\big(U\big)\det\big(P\big)=\pm \det\big(P\big) = \pm(\sigma_1\cdot\sigma_2)$

now I assume you know that $||A||_2 = \sigma_1$ and $||A^{-1}||_2=\frac{1}{\sigma_2}$ so we may write

$\text{cond}_2(A)=||A||_2 \cdot ||A^{-1}||_2 = \frac{\sigma_1}{\sigma_2}=\frac{\sigma_1^2}{\sigma_1\sigma_2}\lt \frac{\sigma_1^2+\sigma_2^2}{\sigma_1\sigma_2}=\frac{\big \Vert A\big \Vert_F^2}{\det\big(A\big)}= \frac{3+a^4}{1-a^2}\lt \frac{4}{1-a^2}$

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