I Eigen Vectors, Geometric Multiplicities and more....

[Bullington]

My professor states that "A matrix is diagonalizable if and only if the sum of the geometric multiplicities of the eigen values equals the size of the matrix". I have to prove this and proofs are my biggest weakness; but, I understand that geometric multiplicites means the dimensions of the solution space for the equation Ax=λx (right?). But what does the "sum of the geometric multiplicities" mean? Could you point me in the right direction, thanks!

 

[BvU]

(right?)

geometric multiplicity is the dimension of the solution space of $\vec{\vec A}\vec x = \lambda_i \vec x$ for one $\lambda_i$. add them up for all $i$ and you get the sum of the geometric multiplicities, which you are asked to prove is equal to the size of A.

 

[Bullington]

How could I add up the dimensions? So for a 3x3 matrix that has three unique eigen vectors would I say that the dimension each of the eigen spaces is 3 and the sum of the geometric multiplicities is 3? Then would I say that A would have to be a square matrix of order 3?

 

[Bullington]

Is this in the right direction:
In order for a matrix “A” to be diagonalizable then there is an equation such that P^-1AP=D where D is the diagonalized matrix and P is the matrix formed from the Eigenvectors of A and if the sum of the geometric multiplicities is less than the size of A then P will not be invertible? Am I too far off, or did I assume something I shouldn't have?

 

[BvU]

How could I add up the dimensions? So for a 3x3 matrix that has three unique eigen vectors would I say that the dimension each of the eigen spaces is 3 and the sum of the geometric multiplicities is 3? Then would I say that A would have to be a square matrix of order 3?

No, if an eigenvector $\vec x$ has a unique eigenvalue $\lambda_x$, all multiples of that vector have the property $\vec{\vec A}(p \vec x) = \lambda_x (p\vec x)\$ (p a real number) so the dimension of the solution space is 1. Three unique eigenvalues let that add up to 3.

If two vectors $\vec x$ and $\vec y$ have the same $\lambda$, then $p\vec x + q\vec y$ has that too and the solution space for that degenerate eigenvalue has dimension 2. One other plus these 2 adds up to 3.

 

[geoffrey159]

=> A is diagonalizable : $A \sim \begin{pmatrix}\lambda_1 \text{ Id}_{m_1} & 0 & 0 \\ 0 & \ddots & 0\\ 0 & 0 & \lambda_p \text{ Id}_{m_p} \end{pmatrix}$. What is $m_1,...,m_p$ ? What is $m_1 + ... + m_p$ equal to ?

<= Say that matrix A represents an endomorphism on vector space $E$. You are given that $\text{dim}(E_{\lambda_1}) + ... + \text{dim}(E_{\lambda_p}) = \text{dim}(E)$. Can you show that $E=E_{\lambda_1} \bigoplus ... \bigoplus E_{\lambda_p}$ ? How does this prove that their exists a basis of $E$ in which A is diagonal ?