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

Kelvin Gregory 2022-02-23 Answered
Here is the example I encountered :
A matrix M(5×5) is given and its minimal polynomial is determined to be (x2)3. So considering the two possible sets of elementary divisors
{(x2)3,(x2)2}  and  {(x2)3,(x2),(x2)}
we get two possible Jordan Canonical forms of the matrix , namely J1 and J2 respectively. So J1 has 2 and J2 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 J1 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 J1 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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Expert Answer

Brody Buckley
Answered 2022-02-24 Author has 5 answers
Let A be a matrix reduced in Jordan form. Notice that for every Jordan block relative to the eigenvalue α there is an eigenvector with respect to α 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 AαI for every eigenvalue α and calculate its rank.
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