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Thanks,

David

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- Thread starter 0pt618
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Thanks,

David

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Is this a homework question??

What did you try alrready??

What did you try alrready??

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Deveno

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HallsofIvy

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[tex]\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]

A matrix is not invertible if and only if it has 0 as an eigenvalue. A matrix is diagonalizable if the exist a basis for the space consisting of eigenvectors. Those are not contradictory.

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[tex]\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]

A matrix is not invertible if and only if it has 0 as an eigenvalue. A matrix is diagonalizable if the exist a basis for the space consisting of eigenvectors. Those are not contradictory.

Uuh, that is already a diagonal matrix...

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HallsofIvy

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If you want a non-diagonal, diagonalizable, matrix that is not invertible, start with a diagonal matrix, say

[tex]D= \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{bmatrix}[/tex]

and multiply by A and [itex]A^{-1}[/itex] where A is some invertible matrix.

For example, if

[tex]A= \begin{bmatrix}1 & 0 & 0 \\ 2 & -1 & 0 \\ 5 & -2 & 1\end{bmatrix}[/tex]

then

[tex]A^{-1}= \begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 2 & 1\end{bmatrix}[/tex]

and then

[tex]ADA^{-1}= \begin{bmatrix}-3 & 2 & 0 \\ -2 & -2 & 0 \\ 13 & -4 & 0\end{bmatrix}[/tex]

which is a non-invertible matrix which can be "diagonalized" to the original matrix, D.

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Yes. So? A diagonal matrix is trivially "diagonalizable".

It's fine. Except that the OP want a matrix that is

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for example:

01

00

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HallsofIvy

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Okay, lets start with a Jordan Normal form, non-diagonal matrix:

[tex]P= \begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0\end{bmatrix}[/tex]

which also has 0 as an eigenvalue and so is not invertible. Using the same "A" as before,

[tex]APA^{-1}= \begin{bmatrix}0 & 1 & 0 \\ 4 & 0 & 0 \\ 8 & 1 & 0\end{bmatrix}[/tex]

That is neither invertible nor diagonalizable.

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