Question 1.11.4: the matrix A from example 1.10.4 in section 1.10, Show that ...

In example 1.10.4, we saw that the eigenvalues of A are λ=−4 (with multiplicity 1) and λ = 3 (with multiplicity 2). In addition, the corresponding eigenvectors are v = [−2   \frac {8}{3}   1]^{T} for λ=−4 and v = [5   −2   1]^{T} for λ = 3. In particular, recall that every scalar multiple of v_{λ}=−4 is also an eigenvector of A corresponding to λ=−4. We now show that the set of all these eigenvectors corresponding to λ=−4 is a subspace of R^{3}.

Let E_{λ}=−4 denote the set of all vectors v such that Av=−4v. First, certainly it is the case that A0 = −40. This shows that the zero element of R^{3} is an element of E_{λ}=−4. Furthermore, we have already seen that every scalar multiple of an eigenvector is itself an eigenvector, and thus E_{λ}=−4 is closed under scalar multiplication. Finally, suppose we have two vectors x and y such that Ax=−4x and Ay =−4y. Observe that by properties of linearity,

A(x +y) = Ax +Ay
=−4x −4y
=−4(x +y)

which shows that (x +y) is also an eigenvector of A corresponding to λ=−4.
Therefore, E_{λ}=−4 is closed under addition.
This shows that E_{λ}=−4 is indeed a subspace of R^{3}. In a similar fashion, E_{λ}=3 is also a subspace of R^{3}.