Give an example of a non-diagonalizable 4x4 matrix with eigenvalues: -1, -1, 1, 1.
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A matrix is diagonalizable if it has a full set of eigenvectors. Every distinct eigenvalue has an eigenvector. A double (repeated) eigenvalue, might not have two eigenvectors. So, we are looking for a matrix that is "missing" some of its eigenvectors.
Consider the matrix: `[[3,0],[0,3]]` .
`| [[3 - lambda ,0],[0,3 - lambda ]] | = (3-lambda)^2 - 0 = 0`
Thus `(3- lambda)^2 = 0` and `lambda =3`
We have two eigenvalues (because our `lambda` equation was quadratic), but both of them are 3. (That is, the "multiplicity" of 3 is 2)
To find the eigenvectors we consider the nullspace of:
`[[3 - lambda ,0],[0,3 - lambda ]] = [[3 - 3 ,0],[0,3 - 3]] = [[0 ,0],[0,0]] `
But this is the whole space, so the eigenvectors are `[[1],[0]]` and `[[0],[1]]`
But what if we modify this slightly?
Consider the matrix
`[[3,1],[0,3]]`
Again,
`| [[3 - lambda ,0],[0,3 - lambda ]] | = (3-lambda)^2 - 1*0 = 0`
Thus `(3- lambda)^2 = 0` and `lambda =3`
So again, we have two eigenvalues, both are 3.
Now, to find the eigenvectors we consider the nullspace of:
`[[3 - lambda ,1],[0,3 - lambda ]] = [[3 - 3 ,1],[0,3 - 3]] = [[0 ,1],[0,0]] `
Now the nullspace is just `[[1],[0]]`, so even though we had two eigenvalues (both 3), we got just one eigenvector.
We can play this game with your eigenvalues:
The matrix
`[[1,0,0,0],[0,1,0,0],[0,0,-1,0],[0,0,0,-1]]`
has a full set of eigenvectors, but the matrix
`[[1,1,0,0],[0,1,0,0],[0,0,-1,0],[0,0,0,-1]]`
has only three eigenvectors, and the matrix
`[[1,1,0,0],[0,1,0,0],[0,0,-1,1],[0,0,0,-1]]`
has only two eigenvectors. Thus, neither of these is diagonalizable.
(Note: we are using the Jordan normal form, if you know what that is. If not, don't worry about it. Basically if a matrix can be diagonalized, the diagonal matrix is the Jordan normal form. If it can't be diagonalized, the Jordan normal form is the "best you can do" or the "closest to diagonal" that you can get.)
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