I tried to answer the following question but stuck in number three. \begin{pmatrix} -1 & 1 & 0 \\ 0 & 2 & 2\\ 0 & 1& 1 \end{pmatrix} 1. find the eigenvalue
$\lambda=3, \lambda=-1,\lambda=0$
2. find eigenvector
when $\lambda=0$
$v_1= \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$

when $\lambda=3$
$v_2= \begin{pmatrix} \frac{1}{2} \\ 2 \\ 1 \end{pmatrix}$
when $\lambda=-1$
$v_3= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$
3. find matrix $P$ that diagonalize $A$, $P^{-1}AP$
here I was confused, although I got all three eigen vector and geometric multiplicity for all eigen value same as algebraic multiplicity but the matrix $A$ itself has $\lambda=0$ as one of its eigen value which mean that the matrix $A$ dont have inverse, and $\det |A|=0$, rank $A$ is $2$, which mean $A$ is linear dependent. So is it mean that there isn't matrix $P$ that can diagonalize it? Is there any relationship between eigen value $\lambda=0$ and invertible and diagonalizable? and is my approach true? if this true should i use gram schmidt to find the ortogonal matrix to diagonalize it? or should i just find the inverse of P? thanks!!


Indeed, the matrix $A$ has no inverse. Nobody claimed it has. Just take$$P=\begin{pmatrix}1&\frac12&1\\1&2&0\\-1&1&0\end{pmatrix}.$$Note that the columns of $P$ are your eigenvectors (disclaimer: I did not check your computations). Then$$P^{-1}AP=\begin{pmatrix}0&0&0\\0&3&0\\0&0&-1\end{pmatrix}.$$

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    $\begingroup$ thanks! so there is no connection between a matrix being invertible, linear independent, and diagonalizable? as long as geometric multiplicity same as algebraic multiplicity and A has n linear independent eigen vector? and for the diagonalization part, if i try to proof that the columns of P are eigenvectors, so i need to find the $P^-1$ or find orthogonal vector using gram schmidt than using $P^TAP$? $\endgroup$ – fiksx Dec 26 '17 at 12:45
  • $\begingroup$ Right, except that I have never heard about matices which are linear dependente or linear independent. But forget Gram-Schmidt. You can't bu sure that you will find an aorthonormal basis of eigenvectors. $\endgroup$ – José Carlos Santos Dec 26 '17 at 14:02

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