Linear Algebra Terminology Trouble I have been reading up on how to bring a matrix to diagonal form, and I learned that a matrix is diagonalizable if and only if the eigenvectors "span the space." What does it mean for eigenvectors to "span the space"? I am inferring  that it means to "fill in" the the columns of a matrix $\textbf{S}^{-1}$ that diagonalizes $\textbf{T}$  using the eigenvectors of $\textbf{T}$. For example, if a matrix $\textbf{T}$  has two colums, it will need two linearly independent eigenvectors to "span the space." Thank you. 
 A: We say that a set of vectors $\{v_1,\ldots, v_n\}\in V$ spans the finite dimensional vector space $V$ if every vector $w\in V$ can be written in the form 
$$\sum_{i=1}^n a_iv_i=w $$
where here $a_1,\ldots, a_n$ are elements of the coefficient field $\mathbb{F}$. $\mathbb{F}$ is most likely $\mathbb{R}$ or $\mathbb{C}$ for your purposes. Basically, the notion of a spanning set boils down to being able to express any vector in the space in terms of combinations of the vectors in the set. For instance, the vector $1$ spans $\mathbb{R}$, because any $k\in \mathbb{R}$ satisfies $k=k*1$. Similarly, $(1,1), (1,0),(0,1)$ spans $\mathbb{R}^2$ because any $(a,b)\in \mathbb{R}^2$ can be written as a combination of $(1,1),(1,0),(0,1)$, as you can check.
A: I think it's probably a convoluted way of saying that the matrix has a full set of eigenvectors. Let's assume that we are talking about real numbers $\mathbb{R}$ and an $n\times n$ matrix.
Informally speaking, two things can wrong when looking for eigenvalues and eigenvectors: 1. Some of eigenvalues could be complex and 2. Some of the eigenvalues could be "defective". Defective means that the algebraic multiplicity is higher than the geometric multiplicity. (The relevant discussion comes 4 lessons after this one: http://lem.ma/JZ)
In both of these "bad" cases, the matrix is not cleanly diagonalizable. In the case of complex eigenvalues, you will end up with $2\times 2$ blocks on the diagonal, and in the defective case you will end up with Jordan blocks, e.g. like this $$\begin{bmatrix}1 & 2 \\ 0&1 \end{bmatrix}$$
Both cases are characterized by having fewer than $n$ linearly independent real eigenvectors. So you have fewer (linearly independent) eigenvectors than the dimension of the space. And because there are too few of them, they don't span the space.
In the opposite case, when all eigenvalues are real and none are defective, the matrix is diagonalizable and you have $n$ (linearly independent) eigenvectors. To sum up:
Diagonalizable = eigenvectors span the space
Nondiagonalizble = eigenvectors fail to span the space
