Determining Eigenvalue 4 Invertibility in Linear Algebra

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To determine if 4 is an eigenvalue of matrix A, one must check the invertibility of A-4I. If A-4I is invertible, then its columns are linearly independent, indicating that the only solution to A-4I=0 is the trivial solution, meaning 4 is not an eigenvalue of A. The definition of an eigenvalue states that it exists if the equation Av=λv has non-trivial solutions. This implies that if the matrix M is invertible, the equation Mv=u has a unique solution. Therefore, the invertibility of A-4I directly impacts the determination of eigenvalue 4.
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To find out if 4 in an eigenvalue of A, decide if A-4I is invertible...

So, if A-4I is invertible, then its cols are lin ind by IMT, and also there is only the trivial solution to A-4I=0, so thus 4 is not an eigenvalue of A

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? The definition of "eigenvalue" is that \lambda is an eigenvalue if and only if Av= \lambda v has non-trivial solutions. That is the same as saying that Av- \lambda v= (A- \lambda I)v= 0 has non-trivial solutions. Since v=0 is obviously a solution, saying it has non-trivial solutions means it does NOT have a "unique" solution. If the matrix M has an inverse, then the equation Mv= u has the unique solution [math]v= M^{-1}u[/math].
 
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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