Why do eigenvectors exist if the matrix is invertible?

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DiamondV
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Homework Statement


Find the eigenvalues and associated eigenvector of the following matrix:
A 3 by 3 matrix with rows (1, 2, -1), (1, 0, 1) and (4, -4, 5), written in large parentheses.


Homework Equations

The Attempt at a Solution


Handwritten working: matrix A with rows (1, 2, -1), (1, 0, 1), (4, -4, 5); column operations c2-2c1 and c3+c1 give (1, 0, 0), (1, -2, 2), (4, -12, 9); det(A) found by expansion along top row as 1(det[[-2, 2], [-12, 9]]) = (-2)(9) - (-12)(2) = 6.


We have a theorem in our lectures notes that states that if a matrix is invertible the only eigenvector in its kernel will be the zero vector. In order to find out if it is invertible we get the det(A) and see if its equal to 0 or not, if it is equal to 0(you can't divide by 0) then there is no inverse, if it is not equal to 0(like in this case I got 6) then it is invertible and the only vector is the zero vector in the kernel. So technically I should stop my calculations at this point and say the zero vector is the only one.
However in the solutions given to use they have an answer that is not a 0 vector.
Three column vectors with entries 1, -1, -2 labelled for 1; 2, -1, -4 labelled for 2; and 1, -1, -4 labelled for 3.

1,2,3 are eigenvalues.
How is this possible?
 
DiamondV said:

Homework Statement


Find the eigenvalues and associated eigenvector of the following matrix:
A 3 by 3 matrix written in parentheses with entries: first row 1, 2, -1; second row 1, 0, 1; third row 4, -4, 5.


Homework Equations

The Attempt at a Solution


Handwritten working on lined paper: matrix A with rows (1, 2, -1), (1, 0, 1), (4, -4, 5); row and column operations c2-2c1 and c3+c1 give a matrix with rows (1, 0, 0), (1, -2, 2), (4, -12, 9); determinant by expansion along top row gives 1 times det of [[-2, 2], [-12, 9]] = (-2)(9) - (-12)(2) = 6.


We have a theorem in our lectures notes that states that if a matrix is invertible the only eigenvector in its kernel will be the zero vector. In order to find out if it is invertible we get the det(A) and see if its equal to 0 or not, if it is equal to 0(you can't divide by 0) then there is no inverse, if it is not equal to 0(like in this case I got 6) then it is invertible and the only vector is the zero vector in the kernel. So technically I should stop my calculations at this point and say the zero vector is the only one.
However in the solutions given to use they have an answer that is not a 0 vector.
Three column vectors with labels: (1, -1, -2) for 1, (2, -1, -4) for 2, and (1, -1, -4) for 3.

1,2,3 are eigenvalues.
How is this possible?

There is no such theorem as the one you state from your lectures. The zero vector is not considered as an eigenvector at all.

There is a theorem stating that if a matrix is invertible (has non-zero determinant) then the only vector (NOT EIGENVECTOR!) in the kernel is the zero vector. That has nothing at all to do with eigenvalues and eigenvectors.

Do you actually know what eigenvalues and eigenvectors are?
 
Ray Vickson said:
There is no such theorem as the one you state from your lectures. The zero vector is not considered as an eigenvector at all.

There is a theorem stating that if a matrix is invertible (has non-zero determinant) then the only vector (NOT EIGENVECTOR!) in the kernel is the zero vector. That has nothing at all to do with eigenvalues and eigenvectors.

Do you actually know what eigenvalues and eigenvectors are?

Oh. Not really. Our lecture notes haven't shown any graphs with vectors on them or any sort of visualisation for this. I just know that there's these things called eigenvectors and eigenvalues that are really useful for some reason.