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

MATH221 December 2009

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[hide] 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 $P^{-1}$.)

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[show] Hint

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[show] Solution

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 $PDP^{-1}={\begin{bmatrix}1&0&2\\1&1&1\\0&1&-1\end{bmatrix}}=A$