Conditions for diagonalizable matrix

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A 3×3 matrix A is diagonalizable if it has three linearly independent eigenvectors, which can be arranged in a non-singular matrix P. The diagonal matrix D can be expressed as D = P_inv * A * P. To show that different eigenvalues yield linearly independent eigenvectors, consider two eigenvectors x and y corresponding to distinct eigenvalues c1 and c2; if a1x + a2y = 0, then it must follow that a1 = a2 = 0. For an n×n matrix with n distinct eigenvalues, the characteristic polynomial has n distinct roots, indicating a full rank and a trivial kernel. Thus, the conditions for diagonalizability hinge on the presence of distinct eigenvalues leading to linear independence among eigenvectors.
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Homework Statement
Show that if a matrix has n distinct eigenvalues then it is diagonalizable.
Relevant Equations
A matrix is diagonalizable if it is similar to a diagonal matrix.
If a 3×3 matrix A produces 3 linearly independent eigenvectors then we can write them columnwise in a matrix P(non singular). Then the matrix D = P_inv*A*P is diagonal.

Now for this I need to show that different eigenvalues of a matrix produce linearly independent eigenvectors.

A*x = c1x
A*y = c2y

c1 !=c2

Then how to show that :

a1x + a2y =0 implies a1=a2=0?
 
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If you have ##n## different eigenvalues, what can you say about the characteristic polynomial and the kernel?
 
I assume you mean to say an nxn matrix with n distinct eigenvalues.
 
Use c1x+c2y=0. Both most be on the same line. Lines deprend on a single parameter, meaning lines through the origin.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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