# Is this matrix diagonalizable?

1. Jan 25, 2006

### stunner5000pt

is htis matrix diagonalizable?
$$\left( \begin{array}{ccc}4&0&0\\0&2&2\\2&3&1 \end{array} \right)$$
the characteristic polynomial is $$(x+1)(x-4)^2$$
whats worng with the multiplicity here?
the back of the book says dim(E4) =1 and here multiplicity is 2?

$$\left( \begin{array}{ccc}3&0&6\\0&-3&0\\5&0&2 \end{array} \right)$$
whose characteristic polynomial is $$(\lambda +3)^2 (\lambda -8) = C_{A} (x)$$
here the multiplicity of lambda = -3 is 2 and the [itex] dim(E_{-3}) = ??
what is the dimension of E (-3)??

2. Jan 25, 2006

### 0rthodontist

The dimension of the eigenspace of -3 is the number of independent eigenvectors you can find for the eigenvalue -3. This will be either 1 or 2 eigenvectors. To find them, row reduce (A - (-3)I)x = 0.

3. Jan 26, 2006

### TD

There's a difference between the algebraic multiplicity (AM, which is the multiplicity of the root in the characteristic polynomial) and the geometric multiplicity (GM, which is the dimension of the eigenspace belonging to that eigenvalue, so the number of linearly independant eigenvectors).
When AM > GM for a certain eigenvalue (GM > AM isn't possible), then the matrix is not diagonalizable since you won't have enough linearly independant eigenvectors.