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I have this Matrix $A$. I want to diagonalise it so that $D = P^{-1}AP$.

$A =\begin{pmatrix} 6 & -2 & 2 \\ -2 & 6 & 2 \\ 4 & 4 & 4\end{pmatrix}$

I found the Eigenvalues $\lambda_1 = 0 , \lambda_2 = 8, \lambda_3 =8$. Since the Eigenvalues are not distinct, is my Matrix Diagonalisable? Since my Matrix A has an Eigenvalue with Multiciplity 2, this means that 2 Eigen Vectors will be Linearly Dependent, and therefore my Matrix is not Diagonalisable? Is there any simple way to get around this.

If someone have an example of the Diagonalisation of a Matrix when going around this that would be great!

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  • $\begingroup$ My Matrix is not Symmetric, look at lower right corner and upper right corners... $\endgroup$ – Xavier Sep 27 '15 at 0:51
  • $\begingroup$ alright. However, after subtracting $8I,$ the resulting matrix really is rank one. The vectors annihilated by it make a two dimensional space, there are two independent vectors as a basis of that $\endgroup$ – Will Jagy Sep 27 '15 at 0:54
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Despite not being symmetric the matrix is diagonalizable. To flesh out Will's comment, $$ A-8I = \left[ \begin{array}{ccc} -2&-2&2 \\ -2&-2&2 \\ 4&4&-4 \end{array} \right] $$ is clearly rank one. Since it maps into $R^3$ it must have a two-dimensional kernel. Two easy vectors seen to span it are $(1,-1,0)$ and $(1,0,1)$. The trouble with using them to diagonalized $A$ is that they are not orthogonal. However, we can use the Gram-Schmidt procedure to produce two orthogonal vectors spanning Ker$(A)$.

Essentially if $v_i$ are our vectors we subtract the projection of $v_2$ onto $v_1$ from $v_2$ to get an $8$-eigenvector normal to $v_1$. With our example, this gives $v_1 =(1,-1,0)$ and $v_2=(1/2,1/2,1)$. Use these two vectors in addition to the $0$-eigenvector to produce the columns of your matrix $P$.

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