How many basis vectors are there in an eigenspace of dimension k?

Question

If $T:V\to V$is a linear map and we know that $\lambda$ is an eigenvalue of $T$ and the eigenspace of T wrt $\lambda$ has dimension $k$ then does that mean there are $k$ linearly independent eigenvectors of T that form a basis of the eigenspace? or is there only 1 eigenvector?

Answer 1

$k$ linearly independent are required to form a basis of a $k$-dimensional eigenspace. In general, the dimension of a vector space refers to the number of (linearly independent) vectors required to build a basis for that space.

Answer 2

For a k-dimensional eigenspace you need $k$ linearly independent eigenvectors.

Note that an eigenvalue may have more than one linearly independent eigenvectors but an eigenvector can not have more that one eigenvalue.