Eigevectors of an Idempotent Matrix: A^2 = A [Strang P310 6.2.25]

Question

Source: P4 of http://www.minho-kim.com/courses/11sp71007/data/hw05-solution.pdf

$\Large{{1.}}$ I perceive : Because $A\mathbf{a_i = 1a_i}$ for all $1 \le i \le n$, thus $1$ is an eigenvalue of $A$
and $\mathbf{a_i}$ are eigenvectors of $A$ for all $1 \le i \le n$. Yet how does this imply the green:
$E_1 = \text{ eigenspace corresponding to the eigenvalue } 1 = \operatorname{span}(\mathbf{a_1}, \cdots, \mathbf{a_n})$ ?

Since $A = \begin{bmatrix} \mathbf{a_1} & \cdots & \mathbf{a_n} \\ \end{bmatrix}$ where $\mathbf{a_i}$ is a column vector,
I apprehend that $\operatorname{span}(\mathbf{a_1}, \cdots, \mathbf{a_n}) = colspace(A)$.

$\Large{{2.}}$ What's the intuition? Does the geometric interpretation of eigenvectors figure here?

Answer 1

By symmetry between $0$ and $1$, there is a second error in the question: not only should $A=I$ be excluded, but also (for part b) it should have excluded $A=0$, since then $1$ is not en eigenvalue (can you spot the error in the "proof" that $1$ is an eigenvalue?).

The fact that the span of the columns (which is the image of the linear map defined by the matrix) must be all of the eigenspace for$~\lambda=1$ is simply because every eigenvector for $1$ is obviously in the image of the linear map (as it is its own image).

As for intuition, there is not much more to offer than that eigenspaces for different eigenvalues (here $0$ and $1$) are always in a direct sum, and that by rank-nullity the sum fills up the whole space here. This proof is not the most insightful one possible for $A^2=A$ implies $A$ diagonalisable. (But Strang wants to push null and column spaces wherever he can.)

Answer 2

Suppose $A^2=A$.

Then $(I-A)^2 = (I-A)$, and $x = Ax+(I-A)x$ for all $x$. In other words, $\mathbb{F}^n = {\cal R} A + {\cal R}(I-A)$.

It is straightforward to verify that ${\cal R}(I-A) = \ker A$ and $\ker A \cap {\cal R} A = \{0\}$.

Let $v_1,...,v_k$ be a basis for $\ker A$, and $v_{k+1},...,v_n$ be a basis for ${\cal R} A$. Then $v_1,...,v_n$ is a basis for $\mathbb{F}^n$. (Note: It is possible that $\ker A = \{0\}$, or $\ker A = \mathbb{F}^n$, in which case one or other basis is empty. This will result in $A$ being the identity or zero operator, respectively.)

We have $Av_i = 0$ for $i=0,...,k$ and $A v_i = v_i$ for $i=k+1,..,n$, and so $A$ is diagonal in this basis, and the diagonal entries are either zero or one.