Deflation of eigenvalue Could you explain to me the deflation?
For example if we have the matrix  $A=\begin{pmatrix}1 & -0.5 & -1.5\\ -15 & -2.5 & 4.5\\ -15 & -4.5 & 2.5\end{pmatrix}$ how do we apply the deflation of the eigenvalue $\lambda_1=4$ ?
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EDIT :
I have done the following :
We have the eigenvalue $\lambda_1=4$. The corresponding eigenvector is : $v_1=\begin{pmatrix}-1\\ 3 \\ 1\end{pmatrix}$, right?
Since $v_1$ does not have $1$ as the component of largest modulus, we multiply  $v_1$ by a permutation matrix $P$ which interchanges the largest element and the first element: We interchange rows $1$ and $2$ of $A$ : $v_1'=\begin{pmatrix}3 \\-1\\  1\end{pmatrix}$
$$\begin{pmatrix}-15 & -2.5 & 4.5\\1 & -0.5 & -1.5\\  -15 & -4.5 & 2.5\end{pmatrix}$$
We want to transform the vector into the vector $e_1$, so we divide the first row by $3$, add it to the second row and subtract it from the third row : $v_1''=\begin{pmatrix}1 \\0\\  0\end{pmatrix}$
$$\begin{pmatrix}-5 & -5/6 & 1.5\\-4 & -4/3 & 0\\  -10 & -11/3 & 1\end{pmatrix}$$ Since we have interchanged the rows $1$ and $2$, now we have to interchange the columns $1$ and $2$ and so we get the matrix $$\begin{pmatrix} -5/6 & -5 &1.5\\-4/3 & -4 & 0\\   -11/3 & -10 &1\end{pmatrix}$$ Then $\lambda_1=4$ should be an eigenvalue of this matrix with $e_1$ the corresponding eigenvector, but this is not like that. What am I doing wrong?
Is the deflated matrix equal to $$\begin{pmatrix} -4 & 0\\   -10 &1\end{pmatrix}$$ ?
 A: $\require{cancel}$
In fact, the method I know under the name "deflation" does not use at all permutation matrices.
Let us say first that matrix
$$A = \pmatrix{1&-0.5&-1.5\\-15&-2.5&4.5\\-15&-4.5&2.5}$$
has eigenvalues $\{-4,4,1\}$.
You just need to find 3 "ingredients" :

*

*"the" (or "one of the") eigenvalue(s) $\lambda$ with largest modulus. In your case, indeed $\lambda = 4$ (you could have chosen $\lambda = -4$),


*a right eigenvector associated with this $\lambda$, here $R=\pmatrix{-1\\3\\1}$ indeed,


*a left eigenvector associated with $\lambda$, which is an ordinary eigenvector of the transpose matrix $A^T$ ; I have obtained:  $L=\pmatrix{5\\1\\-2}$.
for applying the following deflation "recipe" :
$$A_1:=A-\tfrac{\lambda}{R^TL}RL^T$$
Giving the numerical answer:
$$A_1=\pmatrix{-4&-1.5&0.5\\0&0.5&-1.5\\-10&-3.5&0.5}$$
with eigenvalues $\{-4,0,1\}$ as awaited.
Please note that, in the formula above,  $RL^T$ is a $3 \times 3$ matrix, whereas $R^TL$ is a number, the dot product of vectors $R$ and $L$.
Remark : The deflated matrix (as I "understand" it) remains a $3 \times 3$. It doesn't become a $2 \times 2$ matrix.
Sanity check : Let us establish that vector $R$ has become an eigenvector of $A_1$ with eigenvalue $0$ :
$$A_1R=(A-\tfrac{\lambda}{R^TL}RL^T)R=AR-\tfrac{\lambda}{R^TL}R(L^TR)$$
$$A_1R=\lambda R - \tfrac{\lambda}{\cancel{R^TL}}R(\cancel{L^TR})=0$$
(the cancellation takes place because the dot product is commutative)
A: We have $v_1=\begin{pmatrix}-1\\3\\1\end{pmatrix}$.
Let $P$ be the permutation matrix that swaps the first 2 elements.
Then $Pv_1=v_1'=\begin{pmatrix}3\\-1\\1\end{pmatrix}$ has the element with the largest magnitude first.
Let $R$ be the matrix of row operations that transforms $v_1'$ into $e_1$. That is $Rv_1'=e_1$.
Then $R=\begin{pmatrix}1/3\\1/3 & 1\\-1/3 &&1\end{pmatrix}$.
Let $B=RPAP^{-1}R^{-1}$. Then $B$ is similar to $A$ and:
$$Be_1=RPAP^{-1}R^{-1}e_1=RPAP^{-1}v_1'=RPAv_1=RP\lambda_1 v_1=\lambda_1Rv_1'=\lambda_1e_1$$
So $B$ has $\lambda_1$ as eigenvalue with eigenvector $e_1$.
You have found $RPAP^{-1}=\begin{pmatrix} -5/6 & -5 &1.5\\-4/3 & -4 & 0\\   -11/3 & -10 &1\end{pmatrix}$. We can verify that if we multiply it with $v_1'$ we find $\lambda_1 e_1$.
If we multiply it on the right with $R^{-1}=\begin{pmatrix}3\\-1&1\\1&&1\end{pmatrix}$ then we find $B=\begin{pmatrix}4&-5&3/2\\&-4&0\\&-10&1\end{pmatrix}$.
So $B$ has indeed $e_1$ as an eigenvector for $\lambda_1=4$.
The resulting deflated matrix is $B_1=\begin{pmatrix}-4&0\\-10&1\end{pmatrix}$, which we already had in $RPAP^{-1}$ since $R^{-1}$ affected only the leftmost column. This matches your findings.
We can verify that $B_1$ has indeed the same remaining eigenvalues $-4$ and $1$ as $A$, which are on its diagonal since $B_1$ is a triangular matrix.
