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Matrix Algebra - Jordan Matrix

Let

A=\left[\begin{array}{ccc}2&10&5\\-2&-4&-4\\3&5&6\end{array}\right]

.
Find matrix

T \ni TJT^{-1}=A

.

---
my answer

I have found

\lambda_1=\lambda_2=1,\lambda_3=2

J_A=\left[\begin{array}{ccc}1&1&0\\0&1&0\\0&0&2\end{array}\right]

.
Matrix T consists of eigen vectors of A, hence

T=\left[\begin{array}{ccc}5&2&-5\\5&2&-5\\1&1&-2\end{array}\right]

but it means |T|=0 implies T has no invers. So..... ???? Please help..

Reply 1

Zhen Lin

My guess is the matrix is not diagonalisable and you don't actually have enough eigenvectors to form T. You need to use generalised eigenvectors here.

If you have the Jordan normal form

J = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}

, then the change of basis matrix you need is

T = \begin{pmatrix} \mathbf{u} & \mathbf{v} & \mathbf{w} \end{pmatrix}

, where

A \mathbf{u} = \mathbf{u}

A \mathbf{v} = \mathbf{v} + \mathbf{u}

and

A \mathbf{w} = 2 \mathbf{w}

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