In class on Wednesday I mentioned the following result.
A matrix is diagonalisable if and only if it possesses an eigenbasis, that is, a basis consisting of eigenvectors.
Here is the proof:
Suppose a matrix A (an n by n matrix) admits an eigenbasis. That is, there exists a basis of n eigenvectors for A, call them such that for all i. Form a matrix whose columns are the (that is ). Then S is a change of basis matrix, and moreover, we have
which shows that a diagonal matrix, with diagonal entries the corresponding eigenvalues of A.
Conversely, suppose that A is diagonalizable. Then there exists a change of basis matrix S such that with D diagonal. Now this equation tells us that the n by n matrix AS is equal to the n by n matrix SD. In particular, each row of these matrices must agree. The i-th row of this equation is
which tells us that the matrix S is full of eigenvectors. In particular, the eigenvectors of S form a basis and are thus an eigenbasis.
Here are some questions to think about. We can talk about these in office hours on Monday, or next class. [A hint for 1 and 2 is (as always!) the rank-nullity theorem!]
- In the first part of the proof, we built S with eigenvectors of A. Why could we be sure that S was invertible?
- In the second part of the proof, we concluded by showing that S was full of eigenvectors. Why does it follow that these vectors form a basis?
- Find the eigenvectors for the following matrices, and determine the sum of the dimensions of the eigenspaces [to save you some time – the eigenvalues for both are -2, -2, -3]:
NOTE: These questions are not going to be graded and are in no way part of your grade. They are just questions I think could be helpful to try and understand some of these new ideas.