Science:Math Exam Resources/Courses/MATH221/April 2009/Question 12 (i)

MATH221 April 2009

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[hide] Question 12 (i)
There is no matrix A with eigenvectors $(1,1,1)^{T},(1,0,1)^{T}$, and $(2,1,3)^{T}$ with corresponding eigenvalues 1, -1, 4.

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[show] Hint

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[show] Solution

The answer is false. Let $v_{1}={\begin{pmatrix}1\\1\\1\end{pmatrix}},v_{2}={\begin{pmatrix}1\\0\\1\end{pmatrix}},$ and $v_{3}={\begin{pmatrix}2\\1\\3\end{pmatrix}}$ denote the given vectors. A matrix A has eigenvectors $v_{1},v_{2},v_{3}$ with corresponding respective eigenvalues 1, -1, 4 if and only if $A=P\Lambda P^{-1}$, where $P={\begin{pmatrix}1&1&2\\1&0&1\\1&1&3\end{pmatrix}}$ and $\Lambda ={\begin{pmatrix}1&0&0\\0&-1&0\\0&0&4\end{pmatrix}}.$ Thus the matrix A exists if and only if P is invertible (that is, if and only if $v_{1},v_{2},v_{3}$ are linearly independent). We can calculate $\det P=1\neq 0$. Therefore, P is invertible and the desired matrix A will be given by $A=P\Lambda P^{-1}$. Remark: By computing the inverse (not necessary to answer the question), we can determine the explicit form of A as $A={\begin{pmatrix}-9&2&8\\-3&1&3\\-13&2&12\end{pmatrix}}.$

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