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    A matrix  T represents a linear transformation  T: V \to V with respect to the basis  \{\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3} \} where the matrix is  T = \begin{pmatrix} 4 & -3 & -3 \\ 3 & -2 & -3 \\ -3 & 3 &  4 \end{pmatrix} .
    If  T can be represented by a diagonal matrix  D , find  D and a basis of  V with respect to  D which represents  T . Also find non-singular matrix  P such that  D = P^{-1}AP , where  A is the matrix representing  T with respect to  \{ \mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3} \} .
    If  D= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix} and  P = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & -1 & -1 \end{pmatrix} . How would I go about finding the new basis?
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    The new basis is the set of columns of P. A matrix is diagonal if and only if it is represented w.r.t. a basis of eigenvectors.
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    (Original post by nuodai)
    The new basis is the set of columns of P. A matrix is diagonal if and only if it is represented w.r.t. a basis of eigenvectors.
    Are the answers(the new basis) not necessarily unique? If the eigenvalues have algebraic multiplicity 2 for example and you can find two linearly independent eigenvectors, then the matrix  P would depend on how you write out the eigenvectors?
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    (Original post by JBKProductions)
    Are the answers(the new basis) not necessarily unique? If the eigenvalues have algebraic multiplicity 2 for example and you can find two linearly independent eigenvectors, then the matrix  P would depend on how you write out the eigenvectors?
    The answers are certainly not unique, you can choose any basis of eigenvectors you like. You can try and make it 'more unique' by requiring your basis vectors to be normalized and satisfy some kind of orientation relation -- in fact, in some cases (e.g. Hermitian matrices) you can even take your basis to be orthonormal -- but even then you're still free to change sign and so on.
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    Thanks.
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    (Original post by JBKProductions)
    Are the answers(the new basis) not necessarily unique? If the eigenvalues have algebraic multiplicity 2 for example and you can find two linearly independent eigenvectors, then the matrix  P would depend on how you write out the eigenvectors?
    Even if an eigenvalue has multiplicity 1 then there is plenty of choice: Any non-zero vector in a 1-dimensional space is a basis.

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Updated: April 16, 2012
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