# Here is the example I encountered : A matrix M(5\times 5) is given and its minimal poly

Here is the example I encountered :
A matrix $M\left(5×5\right)$ is given and its minimal polynomial is determined to be ${\left(x-2\right)}^{3}$. So considering the two possible sets of elementary divisors

we get two possible Jordan Canonical forms of the matrix , namely ${J}_{1}$ and ${J}_{2}$ respectively. So ${J}_{1}$ has 2 and ${J}_{2}$ has 3 Jordan Blocks.
Now we are to determine the exact one from these two. From the original matrix M, we determined the Eigen vectors and 2 eigen vectors were linearly independent. So the result is that ${J}_{1}$ is the one .So, to determine the exact one out of all possibilities , we needed two information -
1) the minimal polynomial ,together with 2) the number of linearly independent eigen vectors.
Now this was a question-answer book so not much theoretical explanations are given . From the given result , I assume the number of linearly independent eigen vectors -which is 2 in this case - decided ${J}_{1}$ to be the exact one because it has 2 Jordan Blocks. So the equation
"Number of linearly independent eigen vectors=Number of Jordan Blocks"
must be true for this selection to be correct .
Now this equation is not proved in this book or the text book I have read says nothing of this sort
So, that is my question here : How to prove the equation "Number of linearly independent eigen vectors=Number of Jordan Blocks"?
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Brody Buckley
Let A be a matrix reduced in Jordan form. Notice that for every Jordan block relative to the eigenvalue $\alpha$ there is an eigenvector with respect to $\alpha$ and these vectors ar independent. To see this just look at the first column of a generic block. So Jordan blocks independent eigenvectors. To prove the other inequality, consider the matrix $A-\alpha I$ for every eigenvalue $\alpha$ and calculate its rank.