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"Let A = \begin{pmatrix}-1&-1&-2\\ \:2&2&1\\ \:6&2&6\end{pmatrix}

Find the characteristic polynomial of A.

Find the distinct eigenvalues of A and their respective algebraic and geometric multiplicities."

So I calculated the value of the characteristic polynomial and got x^3 -7x^2 + 16 x -12 or (x - 3)(x - 2)(x - 2) and got the eigenvalues of 2 and 3. I believe these are the correct answers. However, I am not sure how to give the geometric and algebraic multiplicities for the eigenvalues. I thought that for the eigenvalue 3 the multiplicity would have been 1 and for the eigenvalue 2 it would have been 2. But it seems like I need two multiplicity values for each.

Any help?

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The numbers you are quoting are the algebraic multiplicity; i.e., $\lambda$ has algebraic multiplicity $n$ when the factor $(x - \lambda)$ appears exactly $n$ times in the factored characteristic polynomial.

The geometric multiplicity of an eigenvalue $\lambda$ is the dimension of the eigenspace corresponding to $\lambda$. That is, what is the size of the largest set of linearly independent eigenvectors you can create for $\lambda$.

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The algebraic multiplicity of an eigenvalue $\lambda$ of an endomorphism $A$ is its multiplicity as a root of the characteristic polynomial $\chi_A$ of the endomorphism, in other words it is the greatest exponent $m$ such that $(x-\lambda)^m$ divides $\chi_A$.

Its geometric multiplicity is the dimension of the eigenspace $E_\lambda$.

One has the inequalities: $$1\le \text{geometric multiplicity}(\lambda)\le\text{algebraic multiplicity}(\lambda). $$

A necessary and sufficient condition for an endomorphism to be diagonalisable is that all of its eigenvalues have geometric multiplicity equal to its algebraic multiplicity.

This condition is automatically satisfied for eigenvalues which are simple roots of $\chi_A$. There results that an endomorphism with only simple eigenvalues is diagonalisable.

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For $$ A = \pmatrix { 1 & 1 \\ 0 & 1} $$ there is one eigenvalue, namely $1$.

The eigenspace for this is $Ker(A - 1\cdot I)$, which is one-dimensional, spanned by $\pmatrix{1 \\ 0}$. Hence the geometric multiplicity, for this eigenvalue, s 1.

Its algebraic multiplicity (how many times it's a root of the char poly) is 2, however.

Something similar may happen in your example; you'd have to compute to find out.

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The eigenvalues of a matrix $A$ are the roots of the characteristic equation, $p(\lambda)$, of $A$. So if $p(\lambda)=(\lambda-3)(\lambda-2)^2$, then the eigenvalues are the solutions to $p(\lambda)=0$. So, you are correct about the eigenvalues being 2 and 3.

The algebraic multiplicity of an eigenvalue $\lambda$ is the multiplicity of $\lambda$ as a root of the characteristic equation. So the algebraic multiplicity of $2$ is 2 (since it is a double root) and the multiplicity of $3$ is 1 (since it is a single root).

The geometric multiplicity of $\lambda$ is the dimension of the eigenspace of $\lambda$. i.e. the solution set to $Ax=\lambda x$. This value will be at least $1$ and it will be less than or equal to the algebraic multiplicity. For example, the geometric multiplicity of $3$ will be $1$ because its algebraic multiplicity is already $1$. For $2$,

$Ax=2x \implies \left[ {\begin{array}{cc} -1 & -1 & -2\\ 2 & 2 & 1\\ 6 & 2 & 6\\ \end{array} } \right]x= \left[ {\begin{array}{cc} -x_1-x_2-2x_3\\ 2x_1+x_2+x_3\\ 6x_1+2x_2+6x_3\\ \end{array} } \right]= \left[ {\begin{array}{cc} 2x_1\\ 2x_2\\ 2x_3\\ \end{array} } \right]$

You can use row reduction to solve for $x$. If you just want the geometric multiplicity (not the actual eigenspace), then you just need to figure out the dimension of the solution space. This number is equivalent to the number of free variables, or the number rows with only $0$'s when the matrix is in echelon form.

From this calculation, we can see that the solution has the form $ \left[ {\begin{array}{cc} (-1/2)x_3 \\ (-1/2)x_3\\ x_3\\ \end{array} } \right]= x_3\left[ {\begin{array}{cc} -1/2\\ -1/2\\ 1\\ \end{array} } \right]=span(\left[ {\begin{array}{cc} -1/2\\ -1/2\\ 1\\ \end{array} } \right])$. So, the geometric multiplicity of the eigenvalue $3$ is $1$.

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The geometric multiplicity is the dimension of the eigenspace of each eigenvalue and the algebraic multiplicity is the number of times the eigenvalue appears in the factorization of the caracteristic polynomial. In your example the algebraic multiplicity of 3 is 1 and this implies that its geometric multiplicity is also 1. Fir the eigenvalue 2 the algebraic multiplicity is 2 because it appears two times in the factorization. For the algebraic multiplicity of 2 you have to compute the eigenspace dimension with eigenvalue 2.

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