Science:Math Exam Resources/Courses/MATH307/December 2010/Question 01 (c)

MATH307 December 2010

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Question 01 (c)
Let $A={\begin{bmatrix}1&0&0\\0&a&0\\0&0&2\end{bmatrix}}$ where a is a real number. (c) For what values of a (if any) is cond(A) = 1/2? Give a reason.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Break this problem up into cases when $\displaystyle a>1$ and when $\displaystyle a\leq 1$. Compute the norms of A and its inverse in each case and try to show this situation is impossible.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that ${\text{cond}}(A)=\left\|A\right\|\left\|A^{-1}\right\|$ Now, since A is diagonal matrix, it is easy to find $A^{-1}$. $A^{-1}={\begin{bmatrix}1&0&0\\0&{\frac {1}{a}}&0\\0&0&{\frac {1}{2}}\end{bmatrix}}$ If $\left|a\right|>1$, $\left\|A^{-1}\right\|=1$ since $\max \left\{\left|1\right|,\left|{\frac {1}{a}}\right|,\left|{\frac {1}{2}}\right|\right\}=1$. And $\left\|A\right\|\geq 2$ since $\max \left\{\left|1\right|,\left|a\right|,\left|2\right|\right\}\geq 2$. Thus, $\left\|A^{-1}\right\|\left\|A\right\|\geq 2.$ Therefore, it is impossible to get ${\text{cond}}(A)={\frac {1}{2}}$. If $\left|a\right|\leq 1$, $\left\|A\right\|=2$ since $\max \left\{\left|1\right|,\left|a\right|,\left|2\right|\right\}=2$. And $\left\|A^{-1}\right\|\geq 1$ since $\max \left\{\left|1\right|,\left|{\frac {1}{a}}\right|,\left|{\frac {1}{2}}\right|\right\}\geq 1.$ Thus, $\left\|A^{-1}\right\|\left\|A\right\|\geq 2.$ Hence, again it is impossible to get ${\text{cond}}(A)={\frac {1}{2}}$. Therefore, there is no such value a which gives ${\text{cond}}(A)={\frac {1}{2}}$

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