Automatic Control Systems Homework Proof on Linear Algebra

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The discussion focuses on proving the relationship between the geometric multiplicity of an eigenvalue and the rank of a matrix. It states that the geometric multiplicity Qi of an eigenvalue λi for an nxn matrix A can be expressed as Qi = n - rank(A - λiI). Participants emphasize the need for a mathematical proof rather than just verification through examples. The kernel of the matrix A - λiI is highlighted as the space of solutions to the equation (A - λiI)u = 0, while the image represents vectors that do not satisfy this equation. Understanding the relationship between the kernel, image, and rank is crucial for the proof.
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Homework Statement



Show that the geometric multiplicity (denoted by Qi) associated with an eigenvalue(lamda i)

of an nxn matrix A can also be expressed as Qi=n-rank(A-(Idendity Matrix)*(lamda i)).

P.S.:: Lamda i is simply a scaler quantity since it is an eigenvalue where A and I(Idendity Matrix) are matrices.

I could not prove this.I can simply try an example and verify that this is correct but our instructor wants us a matematical proof..Thanks very much for now your valuable help...
 
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Do you know how the multiplicity is related to the dimension of the space of solutions to

(A - \lambda_i I_n)u = 0? (*)

You might want to choose the basis of eigenvectors \{ \mathbf{v}_k \} to compute this. The space of solutions to (*) is called the kernel of the matrix A - \lambda_i I_n. The space of vectors that don't satisfy (*) are called the image of A - \lambda_i I_n. The dimension of the image of a matrix is equal to the rank of the matrix. There is a simple relationship between the dimension of the vector space and the dimensions of the kernel and image of a matrix that you can use.
 

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