Question

5. Suppose A is an n × n matrix, whose entries are all real numbers, that has n distinct real eigenvalues. Explain why R n has a basis consisting of eigenvectors of A. Hint: use the “eigenspaces are independent” lemma stated in class.

6. Unlike the previous problem, let A be a 2 × 2 matrix, whose entries are all real numbers, with only 1 eigenvalue λ. (Note: λ must be real, but don’t worry about why this is true). Could R 2 have a basis consisting of eigenvectors of A? Hint: you already answered this on this homework.

Answer 1

5) Given that A is an nxn matrix, whose entries are all real numbers, that has n distinct real eigenvalues.

Then, the matrix A has n distinct eigenvectors corresponding to the n distinct real eigenvalues, which are linearly independent.

There are n vectors (eigenvectors) which are linearly independent and which can span every vector of Rⁿ.

So, the set of eigenvectors of A can be a basis of Rⁿ.

6) Given that A is a 2x2 matrix, whose entries are all real numbers with only one eigenvalue .

Here is real.

Then there will be one linearly independent eigenvector of the matrix A corresponding to the eigenvalue .

The set of eigenvector cannot span all vectors of R² and the dimension of the set is less than the dimension of R².

Therefore, R² could not have a basis consisting of eigenvectors of A.