[Solution] Define a three by three matrix by
Question: Define a three by three matrix by
\[A=\left[\begin{array}{rrr} -2 & -1 & 1 \\ -15 & 2 & 10 \\ 14 & -6 & -9 \end{array}\right]\]- Show that \(\underline{v}=[3,5,2]^{T}\) is an eigenvector for the matrix \(A\) and determine the corresponding eigenvalue, \(\lambda\). Show that this is the only eigenvalue of \(A\) and determine a basis for the \(\lambda\) -eigenspace.
- Is \(A\) diagonalisable? Give a reason for your answer.
- Let \(\underline{w}=[1,0,2]^{T}\) and \(\underline{u}=[1,1,1]^{T}\). Calculate \((A-\lambda I) \underline{w}\) and \((A-\lambda I) \underline{u}\). Does there exist a vector \(\underline{u}^{\prime}\) such that \((A-\lambda I) \underline{u}^{\prime}=\underline{u} ?\) Justify your assertion.
- Using the previous results determine an invertible three by three matrix \(P\) such that \(T=P^{-1} A P\) is upper triangular and calculate \(T\).
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