Finding eigenvalues by inspection? I need to solve the following problem, In this problem, t



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The eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system.
x1 =2x1+x2x3
x2 =4x13x2x3
x3 =4x1+4x2+2x3
How to find the eigenvalues by using the fact that |AλI|=0, how do it by inspection?

Imagine you have the matrix, A=(211121112)
By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0.

Answer & Explanation

Elois Puryear

Elois Puryear

Beginner2022-01-19Added 30 answers

Let A - the matrix of coefficients, then (Ax)k=λxk

Add the third and the second row. 

So x2+x3=λ(x2+x3), thus λ=1 or x2+x3=0. Since the trace of A equals 1, the sum of other two eigenvalues is 0. 
For other eigenvalues, let x3=x2, giving 2x1+2x2=λx1 and 4x12x2=λx2. The characteristic polynomial is easily found as λ2+4=0.

Terry Ray

Terry Ray

Beginner2022-01-20Added 50 answers

The first eigenvalue may be inferred from the fact that A-I has two equal rows and two equal columns.
The additional two eigenvalues are then provided to you by the trace and determinant.



Expert2022-01-24Added 556 answers

The determinant of a matrix is the sum of the eigenvalues., so in this case λ1λ2λ3=4. The sum of the diagonals is the sum of the eigenvalues so λ1+λ2+λ3=23+2=1[110]T is an eigenvector with eigenvalue 1, then λ1=1 so λ2λ3=4 and λ2+λ3=0. Thus λ2,λ3=±21.In general there is no easy way to spot eigenvalues of 3×3 matrices.With 2×2 matrices you can use the determinant and trace. The fact that these are the only non-trivial coefficients of the characteristic polynomial is the primary cause of why this works (and thus determine the eigenvalues). Other characteristic polynomial coefficients aren't as simple to use or recall when working with larger matrices, which necessitates the employment of more invariants (i.e., coefficients of the characteristic polynomial).

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