# Matrix with repeated eigenvalues is diagonalizable...?

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1. Nov 5, 2015

### kostoglotov

MIT OCW 18.06 Intro to Linear Algebra 4th edt Gilbert Strang

Ch6.2 - the textbook emphasized that "matrices that have repeated eigenvalues are not diagonalizable".

imgur: http://i.imgur.com/Q4pbi33.jpg

and

imgur: http://i.imgur.com/RSOmS2o.jpg

Upon rereading...I do see the possibility to interpret this to mean that fewer than n independent eigenvectors leads to an undiagonalizable matrix...that n-all different eigenvalues ensures n-independent eigenvectors...leaving open the possibility of n-independent eigenvectors with repeated eigenvalues...? Yes, no?

Because the second worked example shows a matrix with eigenvalues 1,5,5,5, and the use of diagonalization of that matrix, and Matlab is quite happy to produce a matrix of n-independent eigenvectors from this matrix.

The matrix in question is 5*eye(4) - ones(4);

What is the actual rule, because I don't feel clear on this.

2. Nov 5, 2015

### rs1n

If all n eigenvalues are distinct, then the matrix is diagonalizable. However, the converse is not true. There are matrices that are diagonalizable even if their eigenvalues are not distinct, as your example clearly shows.

3. Nov 5, 2015

### Staff: Mentor

Diagonalizable means per definition that you can find a basis of eigenvectors. If all eigenvalues exist in the underlying field or ring and there are as many as the dimension is, there is clearly a basis of eigenvectors. This condition is sufficient but not necessary since diag(1,...,1) is diagonalized with just one (repeated) eigenvalue 1. A necessary condition is that the characteristic polynomial det(A-t*id) of a matrix A has only linear factors (which may be repeated, e.g. det(id - t*id) = (1-t)^dimension ). i.e. all zeros of the characteristic polynomial exist in the underlying field or ring. Those zeros are exactly the eigenvalues.

Ps: You have still to find a basis of eigenvectors. The existence of eigenvalues alone isn't sufficient. E.g.
0 1
0 0
is not diagonalizable although the repeated eigenvalue 0 exists and the characteristic po1,0lynomial is t^2. But here only (1,0) is a eigenvector to 0.

Last edited: Nov 5, 2015