I am using the textbook \"Linear Algebra\" third edition byLay. Based on the def

Question

I am using the textbook "Linear Algebra" third edition byLay. Based on the definition on page 303, I understand that thereis ONE eigenvector (ignoring multiplying by ascalar) corresponding ONE eigenvalue. However on page 321-322 example 3, for a 3x3 matrix [1 3 3 -3-5 -3 3 3 1] with eigenvalues of 1 and -2 it says there are 3eigenvectors, v1, v2, and v3. Ax=x is satisfied with =1 and x=v1. Alsosatisfied with = -2 and x= v2+v3. So, are there 2 eigenvectors, v1 and v2+v3? Or are there 3 eigenvectors, v1, v2, and v3? If there are 3, what is the definition of aneigenvector? I am using the textbook "Linear Algebra" third edition byLay. Based on the definition on page 303, I understand that thereis ONE eigenvector (ignoring multiplying by ascalar) corresponding ONE eigenvalue. However on page 321-322 example 3, for a 3x3 matrix [1 3 3 -3-5 -3 3 3 1] with eigenvalues of 1 and -2 it says there are 3eigenvectors, v1, v2, and v3. Ax=x is satisfied with =1 and x=v1. Alsosatisfied with = -2 and x= v2+v3. So, are there 2 eigenvectors, v1 and v2+v3? Or are there 3 eigenvectors, v1, v2, and v3? If there are 3, what is the definition of aneigenvector?

Explanation / Answer

Suppose L is an eigenvalue of the matrix A. Then it's correspondingeigenvector is simply a vector v such that: Av = Lv And that's it. That's just the definition. Note that it is possibleto have more than one eigenvector which each have the sameeigenvalue. If we use your matrix A given above, then theeigenvectors are: v1 = [1,-1,1] with eigenvalue 1 v2 = [-1,0,1] with eigenvalue -2 v3 = [-1,1,0] with eigenvalue -2 Now it's actually the case that if two vectors share the sameeigenvalue, then any linear combination of them will be a vectorwith the same eigenvalue. Note that a linear combination of twovectors u and v is just au + bv, for some scalars a and b. Now in your case, let's test to see whether av2 + bv3 is aneigenvector: A(av2 + bv3) = A(av2) + A(bv3) = a(Av2) + b(Av3) = -2av2 -2bv3 =-2(av2 + bv3) Thus, for any scalars a and b we choose, a*v2 + b*v3 will be a neweigenvector, again with eigenvalue -2. so there are actuallyinfinitely many eigenvectors with eigenvalue -2. ------------------------------------- Note that we can determine whether there will be multipleeigenvectors associated with one eigenvalue by its multiplicity.I'm assuming that you're solving for eigenvalues using thecharacteristic equation. In the characteristic equation for yourmatrix A given above, it should factor to: (L - 1)(L + 2)^2 Since L+2 is squared -2 is called a root of multiplicity 2.This means it should have two distinct eigenvectors associated withit (and any linear combination of those). Hope that helps! Please PM me if you are still unclear aboutthis.

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