Understanding the bases of an eigenspace

Question

Let $A∈ℝ^{2×2}$ be given by$$A=\begin{pmatrix} 1&-2\\ 0&1\\ \end{pmatrix}.$$ Task: Find the bases for the eigenspace.

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So, it is clear that the only distinct eigenvalue is $\lambda=1$. Thus, we can find the eigenvector that forms a bases for the eigenspace corresponding to $\lambda=1$ by solving $(A-\lambda I)\vec{x}=\vec{0}$.

I calculate that this is exactly equivalent to solving (putting in RREF) $$\begin{pmatrix} 0&-2&0\\ 0&0&0\\ \end{pmatrix}.$$ As elementary as this sounds for this level in Linear Algebra, I'm having a hard time actually forming my bases from the solution as a vector. So, I can just rescale right, $$\implies\begin{pmatrix} 0&1&0\\ 0&0&0\\ \end{pmatrix}.$$ So I guess my question is, how do I write the solution set of this matrix, i.e. $$\begin{pmatrix} 0&1\\ 0&0\\ \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \end{pmatrix}.$$

Does the structure of this system imply that $x_2$ is a free variable? Can I just say: $$\begin{pmatrix} 0&0\\ 0&1\\ \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \end{pmatrix}$$ and then take $x_1$ to be my free variable?
Thus the solution set would be $\begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}=\begin{pmatrix} t\\ 1\\ \end{pmatrix}$ where $t\in\mathbb R$, correct? How do I get a singular bases vector out of this last solution if it's correct?

Thanks in advance! P.S. Let me know if there is anyway I can improve my question writing on here, thanks!

Answer 1

You have the constraint $x_2=0$ and $x_1$ is free.

Hence the eigenvectors is $\operatorname{Span} \left\{\begin{pmatrix}1 \\ 0 \end{pmatrix} \right\}$

Answer 2

You think to complicated. Consider $$ \begin{pmatrix} 0&1\\ 0&0\\ \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \end{pmatrix}\Leftrightarrow x_2=0. $$ Hence, the eigenspace is given by all vectors $(x_1,x_2)^T$ such that $x_2=0$. Hence you can write $(x_1,0)^T=x_1(1,0)^T$ for arbitrary $x_1\in\mathbb R$. Finally, you see that all elements of the eigenspace are spanned by $(1,0)^T$ and this forms the base for your eigenspace.

Answer 3

Starting from this: $$\begin{pmatrix} 0&1\\ 0&0\\ \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \end{pmatrix}$$

we get that solutions are given by $$x = \pmatrix{t\\0}$$

To see that this is true, have a look at row 1: $$0\cdot x_1+1\cdot x_2\overset{!}{=}0$$ Here you see that you can plug in any value of $x_1$. But since that equation reduces to $x_2\overset{!}{=}0$, $x_2$ has to be $0$.

Your choice $(t,1)$ does not solve the equation, since $$0\cdot t+1\cdot 1\neq0$$

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The eigenspace consists of all vectors that solve the given equation system. Hence:

$$\text{Eig}(A;1)=\{\pmatrix{t\\0}, t∈ℝ\}$$ and a basis of this space can be given simply by choosing a (non-trivial) vector, e.g. $v_1=(1,0)$.