Science:Math Exam Resources/Courses/MATH221/December 2009/Question 10 (a)

MATH221 December 2009

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Question 10 (a)
Find the number a which make the matrix below diagonalizable. (No need to diagonalize them.) ${\begin{bmatrix}0&1&2\\0&3&a\\0&0&0\end{bmatrix}}$

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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 an n by n matrix to be diagonalizable, the matrix must have n distinct eigenvectors. First let’s find the eigenvalues of a ${\text{det}}({\begin{bmatrix}0-\lambda &1&2\\0&3-\lambda &a\\0&0&0-\lambda \end{bmatrix}})=\lambda ^{2}(3-\lambda )=0$ Solve the equation above to get $\displaystyle \lambda _{1}=3,\lambda _{2}=\lambda _{3}=0$ We got a double root here, so we need to have 2 corresponding eigenvectors for $\lambda =0$. When $\displaystyle \lambda =0,A-\lambda I={\begin{bmatrix}0&1&2\\0&3&a\\0&0&0\end{bmatrix}}$ In order to have 2 eigenvectors, we need to have 2 free variables here i.e. we need to have ${\frac {1}{3}}={\frac {2}{a}}$ Solve this to get $a=6$.

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