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

MATH307 December 2010

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Question 01 (a)
Let $A={\begin{bmatrix}1&0&0\\0&a&0\\0&0&2\end{bmatrix}}$ where a is a real number. (a) For what values of a (if any) does the norm have the value $\|A\|=2$?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
In this special case, the matrix norm of a diagonal matrix with diagonal entries $\displaystyle \lambda _{1},\lambda _{2},.....,\lambda _{n}$ can be easily computed.

Hint 2
It is also possible to directly use the definition of the matrix norm, which is given by $\parallel A\parallel =\max _{{\vec {x}},\parallel {\vec {x}}\parallel =1}\parallel A{\vec {x}}\parallel$

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Solution 1
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since the matrix norm of a diagonal matrix with diagonal entries $\displaystyle \lambda _{1},\lambda _{2},.....,\lambda _{n}$ is the largest value of $\left|\lambda _{k}\right|$, if $\left|a\right|\leq 2$, then $\left\|A\right\|=2$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The matrix norm is given by $\parallel A\parallel =\max _{{\vec {x}},\parallel {\vec {x}}\parallel =1}\parallel A{\vec {x}}\parallel$ Let ${\vec {x}}={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}$, where $\parallel {\vec {x}}\parallel =1$ For $A={\begin{bmatrix}1&0&0\\0&a&0\\0&0&2\end{bmatrix}}$ $A{\vec {x}}={\begin{pmatrix}x_{1}\\ax_{2}\\2x_{3}\end{pmatrix}}$ So, ${\begin{aligned}\parallel A{\vec {x}}\parallel &=(x_{1}^{2}+a^{2}x_{2}^{2}+4x_{3}^{2})^{0.5}\leqslant (\max\{1,a^{2},4\}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}))^{0.5}\\&=\max\{1,|a|,2\}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{0.5}\\&=\max\{1,|a|,2\}\end{aligned}}$ Notice that each of these values is obtained by taking our vector x to be one of the unit basis vectors. Thus for any unit vector ${\vec {x}}$ we have established that $\parallel A\parallel =\max\{1,|a|,2\}$. Going by this, $\parallel A\parallel =2\Leftrightarrow |a|\leqslant 2$ The vector ${\vec {x}}$ that will maximize the magnitude of $||A{\vec {x}}||$ in this example is ${\begin{pmatrix}0\\0\\1\end{pmatrix}}$ if $\displaystyle |a|\leq 2$

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