Science:Math Exam Resources/Courses/MATH221/December 2009/Question 06

MATH221 December 2009

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Question 06
Let A = ${\begin{bmatrix}1&0&2\\1&1&1\\0&1&-1\end{bmatrix}}$. Find an invertible matrix P and a diagonal matrix D, such that $A=PDP^{-1}$. (No need to find $P^{-1}$.)

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 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

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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. The diagonal matrix $D$ will be the eigenvalues of $A$, such that: $D={\begin{bmatrix}\lambda _{1}&0&0\\0&\lambda _{2}&0\\0&0&\lambda _{3}\end{bmatrix}}$ where $\lambda _{i}$ are the eigenvalues, and $\lambda _{1}\geq \lambda _{2}\geq \lambda _{3}$. The invertible matrix P will be the corresponding eigenvectors, such that: $P={\begin{bmatrix}|&|&|\\{\vec {v_{1}}}&{\vec {v_{2}}}&{\vec {v_{3}}}\\|&|&|\end{bmatrix}}$ where ${\vec {v_{i}}}$ is the eigenvector corresponding to the eigenvalue $\lambda _{i}$. The first thing that we need to do is to determine the eigenvalues of $A$. To do this, we take: $\mathrm {det} (A-\lambda I)=0$. From this, we get: ${\begin{vmatrix}1-\lambda &0&2\\1&1-\lambda &1\\0&1&-1-\lambda \end{vmatrix}}=0$ Computing the determinant, we get that $\lambda (\lambda -2)(\lambda +1)=0$ and therefore the eigenvalues are: $\displaystyle \lambda _{1}=2,\lambda _{2}=0,\lambda _{3}=-1$ Next, we must compute the eigenvectors corresponding to these eigenvalues. To do this we must solve: $A{\vec {v}}=\lambda {\vec {v}}$ for each eigenvalue/eigenvector combination. For $\lambda _{1}$: ${\begin{bmatrix}1&0&2\\1&1&1\\0&1&-1\end{bmatrix}}{\vec {v_{1}}}=2*{\vec {v_{1}}}$ Solving this, we get that ${\vec {v_{1}}}={\begin{bmatrix}2\\3\\1\end{bmatrix}}$ For $\lambda _{2}$: ${\begin{bmatrix}1&0&2\\1&1&1\\0&1&-1\end{bmatrix}}{\vec {v_{2}}}=0*{\vec {v_{2}}}$ Solving this, we get that ${\vec {v_{2}}}={\begin{bmatrix}-2\\1\\1\end{bmatrix}}$ For $\lambda _{3}$: ${\begin{bmatrix}1&0&2\\1&1&1\\0&1&-1\end{bmatrix}}{\vec {v_{3}}}=-1*{\vec {v_{3}}}$ Solving this, we get that ${\vec {v_{3}}}={\begin{bmatrix}1\\0\\-1\end{bmatrix}}$ Now that we have the eigenvalues and eigenvectors, we can put these into matrices to get $P$ and $D$! $P={\begin{bmatrix}2&-2&1\\3&1&0\\1&1&-1\end{bmatrix}}$ $D={\begin{bmatrix}2&0&0\\0&0&0\\0&0&-1\end{bmatrix}}$ To check that these matrices are correct, we can compute $PDP^{-1}$. When we do this calculation, we see that $PDP^{-1}={\begin{bmatrix}1&0&2\\1&1&1\\0&1&-1\end{bmatrix}}=A$