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For most of my answers for question 5, I got points off bc the grader said my eigenvector was incorrect, yielding an incorrect general solution.

for A2 in question 5 for example, my third eigenvector was [0;-1;0], which makes sense to me because it's actually an eigenvector of the eigenvalue '-2.' The difference is that I didn't have to multiply by 't' in most of my solutions, but I believe they're still correct. Can you explain why this approach is incorrect?

Thanks,
Chuck

ctalley1 wrote:For most of my answers for question 5, I got points off bc the grader said my eigenvector was incorrect, yielding an incorrect general solution.

for A2 in question 5 for example, my third eigenvector was [0;-1;0], which makes sense to me because it's actually an eigenvector of the eigenvalue '-2.' The difference is that I didn't have to multiply by 't' in most of my solutions, but I believe they're still correct. Can you explain why this approach is incorrect?

If it's a "regular" eigenvector, meaning it satisfies (A - \lambda I)\di=0 (no powers on the matrix), then there won't be any terms multiplied by t in its part of the solution. If it's a generalized eigenvector, meaning that (A-\lambda I) must be raised to a power greater than one before it is equal to zero when \xi is multiplied into it, then there will be terms multiplied by t.

Professor Goodwine,

For question 5 matrix A5, since A(1,2)=1, wouldn't there be a t*[1; 0; 0; 0; 0] part added to the eigenvector [0; 1; 0; 0; 0;]?

I think there's a similar problem with the solution for A4 as well.

Thanks!

ctalley1 wrote:For most of my answers for question 5, I got points off bc the grader said my eigenvector was incorrect, yielding an incorrect general solution.

for A2 in question 5 for example, my third eigenvector was [0;-1;0], which makes sense to me because it's actually an eigenvector of the eigenvalue '-2.' The difference is that I didn't have to multiply by 't' in most of my solutions, but I believe they're still correct. Can you explain why this approach is incorrect?

I just realized something else about your question. If your third eigenvalue was (0,-1,0), what was the second one? You have to have a full set of linearly independent eigenvalues.

lawnoy wrote:Professor Goodwine,

For question 5 matrix A5, since A(1,2)=1, wouldn't there be a t*[1; 0; 0; 0; 0] part added to the eigenvector [0; 1; 0; 0; 0;]?

I think there's a similar problem with the solution for A4 as well.

You are right.