Science:Math Exam Resources/Courses/MATH221/April 2013/Question 08

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MATH221 April 2013

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Question 08
Let $A=\left[{\begin{array}{ccc}2&0&3\\0&3&0\\4&0&1\end{array}}\right]$ Find a diagonal matrix D and an invertible matrix P such that $A=PDP^{-1}$ . Do not compute the matrix $P^{-1}$ . Show that $A^{-k}$ approaches the zero matrix as k becomes very large.

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
You first need to compute the eigenvalues and eigenvectors for the matrix. How will you then package them together to find P and D?

Hint 2
For the second part, use that $\displaystyle A^{-k}=(PDP^{-1})^{-k}$ Then expand and simplify the right hand side.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

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Solution
To find the diagonal matrix D, we need to compute the determinant of the matrix A-tI. ${\begin{aligned}\det(A-tI)&=\det \left(\left[{\begin{array}{cccc}2-t&0&3\\0&3-t&0\\4&0&1-t\\\end{array}}\right]\right)\\&=(3-t)\det \left(\left[{\begin{array}{cc}2-t&3\\4&1-t\end{array}}\right]\right)\\&=(3-t)\left((2-t)(1-t)-12\right)\\&=(3-t)(t-5)(t+2)\end{aligned}}$ Which means that eigenvalues are 3, 5 and -2. To find the eigenvectors (which will be the columns of the matrix P), we pick vectors of the kernels of $A-tI$ where t = 3, 5, -2. Eigenvalue t = 3. $\left[{\begin{array}{cccc}2-3&0&3\\0&3-3&0\\4&0&1-3\\\end{array}}\right]=\left[{\begin{array}{cccc}-1&0&3\\0&0&0\\4&0&-2\\\end{array}}\right]$ Thus an eigenvector corresponding to the eigenvalue t = 3 is given by $\left[{\begin{array}{c}0\\1\\0\end{array}}\right]$ Eigenvalue t = 5. $\left[{\begin{array}{cccc}2-5&0&3\\0&3-5&0\\4&0&1-5\\\end{array}}\right]=\left[{\begin{array}{cccc}-3&0&3\\0&-2&0\\4&0&-4\\\end{array}}\right]$ Thus an eigenvector corresponding to the eigenvalue t = 5 is given by $\left[{\begin{array}{c}1\\0\\1\end{array}}\right]$ Eigenvalue t = -2. $\left[{\begin{array}{cccc}2-(-2)&0&3\\0&3-(-2)&0\\4&0&1-(-2)\\\end{array}}\right]=\left[{\begin{array}{cccc}4&0&3\\0&5&0\\4&0&3\\\end{array}}\right]$ Thus an eigenvector corresponding to the eigenvalue t = -2 is given by $\left[{\begin{array}{c}3\\0\\-4\end{array}}\right]$ can be chosen as an eigenvector. So the matrices D and P are given by $D=\left[{\begin{array}{ccc}3&0&0\\0&5&0\\0&0&-2\end{array}}\right]\quad P=\left[{\begin{array}{ccc}0&1&3\\1&0&0\\0&1&-4\\\end{array}}\right]$ Now, as $A=PDP^{-1}$ , then ${\begin{aligned}A^{-k}&=\left(PDP^{-1}\right)^{-k}\\&=PD^{-k}P^{-1}\\&=P\left[{\begin{array}{ccc}3&0&0\\0&5&0\\0&0&-2\\\end{array}}\right]^{-k}P^{-1}\\&=P\left[{\begin{array}{ccc}3^{-k}&0&0\\0&5^{-k}&0\\0&0&(-2)^{-k}\\\end{array}}\right]P^{-1}\end{aligned}}$ So we can see that as k gets large, $D^{-k}$ get closers to the zero matrix. Hence the right hand side, and with it the matrix A_-k, also approaches the zeros matrix.