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Let P(lambda) = lambda^n + a(n-1)lambda^n-1 + a(n-2)lambda^n-2 + ...+ a(1)lambda

ID: 2985465 • Letter: L

Question

Let P(lambda) = lambda^n + a(n-1)lambda^n-1 + a(n-2)lambda^n-2 + ...+ a(1)lambda + a(0) be the characterictics polynomial of matrix A.

It can be shown that constant term a(0) = det A

Prove that zero is an eignvalue of A IFF

A is a singular matrix



I know that the matrix is 4x4

I started with

<==

Let k = 0 is an eigenvalue. Notice that if we plug zero into this equation for k, we just get det(A)=0. This means the matrix is singluar.


==>

Let det(A)=0. Then as stated above we need to find solutions of the equation det(A-kI)=0. Notice that k=0 is a solution since det(A-(0)I) = det(A) which we already know is zero. Thus zero is an eigenvalue.


Did my attempted answers anywhere correct


Explanation / Answer

Your answers are totally right. That is a proof. If you have more questions, please comment here on my answer. Please rate :).

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