Science:Math Exam Resources/Courses/MATH307/April 2009/Question 01 (c)

MATH307 April 2009

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Question 01 (c)
Given the matrix $A={\begin{bmatrix}-10&0&0\\0&-3.5&0\\0&0&0.5\end{bmatrix}}$ Determine the matrix norm of A and cond(A).

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?

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Hint

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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. For a diagonal matrix, the matrix norm is just the maximum absolute value of the diagonal entries. The diagonal values on a diagonal matrix are its eigenvalues. The matrix norm for matrix A: $\left|\left|A\right|\right|=max\left\{\left|\lambda _{i}\right|\right\}$ where $\lambda _{i}$ are the eigenvalues of matrix A (its diagonal entries). Therefore, $\left|\left|A\right|\right|=max\left\{\left|-10\right|,\left|-3.5\right|,\left|-0.5\right|\right\}$ $\left|\left|A\right|\right|=max\left\{10,3.5,0.5\right\}$ $\left|\left|A\right|\right|=10$ The condition number ($cond\left(A\right)$) measures the relative stretching ratio of matrix A. The stretching ratio is defined as ${\frac {max\ stretch}{min\ stretch}}$ which then equates to $\left\|A\right\|\left\|A^{-1}\right\|$. $cond\left(A\right)=\left\|A\right\|\left\|A^{-1}\right\|$ One way to get the inverse of matrix A is to write an augmented matrix comparing matrix A to the identity matrix I. $\left[A|I\right]$ $={\begin{bmatrix}10&0&0\ \ \ \ |&1&0&0\\0&3.5&0\ \ \ \ |&0&1&0\\0&0&0.5\ |&0&0&1\end{bmatrix}}$ The next steps are to row reduce until you get the identity matrix on the left side, and then you will get the inverse of matrix A on the right. This is simple for matrix A since it is a diagonal matrix. All you need to do is to divide each row by the entry on its diagonal to make the entry equal to 1. After you complete this, it should look like this: ${\begin{bmatrix}1&0&0\ \ \ \ |&{\frac {1}{10}}&0&0\\0&1&0\ \ \ \ |&0&{\frac {2}{7}}&0\\0&0&1\ \ \ \ |&0&0&2\end{bmatrix}}$ Therefore, the inverse of matrix A is: $A^{-1}={\begin{bmatrix}&{\frac {1}{10}}&0&0\\&0&{\frac {2}{7}}&0\\&0&0&2\end{bmatrix}}$ We get the matrix norm for $A^{-1}$ just like how we got it for A. Since $A^{-1}$ is still a diagonal matrix, we still use for method given above. $\left\|A^{-1}\right\|=max\{|{\frac {1}{10}}|,|{\frac {2}{7}}|,|2|\}$ $\left\|A^{-1}\right\|=2$ Now that we have $\|A\|$ and $\|A^{-1}\|$, we can get the condition number by multiplying them together. $cond\left(A\right)=\left\|A\right\|\left\|A^{-1}\right\|$ $cond\left(A\right)=10*2$ $cond\left(A\right)=20$ FINAL ANSWER $\left|\left|A\right|\right|=10$ $cond\left(A\right)=20$