Find bases for eigenspaces of A

Question

$$A = \begin{pmatrix} 6 & 4 \\ -3 & -1\end{pmatrix}$$ Find the bases for eigenspaces $E_{\lambda_1}$ and $E_{\lambda_2}$ of $A$.

I don't really know where to start on this problem.

Answer 1

First step: find the eigenvalues, via the characteristic polynomial $$\det(A - \lambda I) = \begin{vmatrix} 6 - \lambda & 4 \\ -3 & -1-\lambda \end{vmatrix} = 0 \implies \lambda^2 - 5\lambda + 6 = 0.$$

One of the eigenvalues is $\lambda_1 = 2$. You find the other one.

Second step: to find a basis for $E_{\lambda_1}$, we find vectors $\mathbf{v}$ that satisfy $(A-\lambda_1 I)\mathbf{v} = \mathbf{0}$, in this case, we go for: $$(A-2I)\mathbf{v} = \begin{pmatrix} 4 & 4 \\ -3 & -3 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies \begin{cases} 4v_1 + 4v_2 = 0 \\ -3v_1 -3v_2 = 0 \end{cases} \implies v_1 = -v_2.$$

So, $\mathbf{v} = (v_1,v_2) = (v_1,-v_1) = v_1(1,-1)$, so $(1,-1)$ is a basis for that eigenspace with eigenvalue $\lambda_1$. Try to find a basis for the other one.

Answer 2

Consider the matrix $$\lambda I - A = \begin{pmatrix}\lambda-6 & -4 \\ 3 & \lambda +1\end{pmatrix}$$ Now, our eiegnavlues of $A$, are the solution to the equation $\det{(\lambda I - A)}=0$ \begin{align}\det{(\lambda I - A)}&=0 \\ \implies(\lambda -6)(\lambda+1)-(3)(-4) &=0 \\ \implies \lambda^2-5\lambda +6 &= 0 \\\implies(\lambda-2)(\lambda-3) &=0 \\ \implies \lambda_1=2 \text{ and } \lambda_2=3\end{align} Thus $\lambda_1=2$ and $\lambda_2=3$ are the two eigenvalues corresponding to matrix $A$.

Since we have two distinct eigenvalues, we know that we will have two eigenvectors of $A$, $\bar{x}= \begin{pmatrix}x_1 \\ x_2\end{pmatrix}$, that corresponds to each eigenvector.

To find these eigenvectors, we must have that is satisfies the equation $(\lambda I - A)\bar{x}=\bar{0}$

For $\lambda_1 =2$: $$(2I-A)=\begin{pmatrix}-4 & -4 \\ 3 &3\end{pmatrix}$$ Which, reduces to $$\begin{pmatrix}1 &1 \\ 0 &0\end{pmatrix}$$ Thus we know $x_1 +x_2 = 0$. Let $x_2=t, \ t\in \mathbb{R} \implies x_1= -t$

Thus $\bar{x}= \begin{pmatrix}x_1 \\ x_2\end{pmatrix} = t \begin{pmatrix}-1 \\ 1\end{pmatrix}, t \in \mathbb{R}$ is the eigenvector of $A$ corresponding to the eigenvalue $\lambda_1 =2$.

Thus the basis for the eigenspace of $A$ corresponding to $\lambda_1 = 2$, is given by $$E_{\lambda_1}=\bigg \{ \begin{pmatrix} -1 \\ 1\end{pmatrix} \bigg \}$$

For $\lambda_2 = 3$:

$$(3I - A) = \begin{pmatrix}-3 & -4 \\ 3 & 4\end{pmatrix}$$ which reduces to $$\begin{pmatrix}1 & \frac{4}{3} \\ 0 & 0\end{pmatrix}$$ Thus we know $x_1 + \frac{4}{3}x_2 =0$. Let $x_2 = s, s\in \mathbb{R} \implies x_1 = -\frac{4}{3}s$

Thus $\bar{x}= \begin{pmatrix}x_1 \\ x_2\end{pmatrix} = s\begin{pmatrix}-\frac{4}{3} \\ 1\end{pmatrix}$ is the eigenvector of $A$ corresponding to the eigenvector $\lambda_2=3$. Thus the basis for the eigenspace of $A$ corresponding to eigenvalue $\lambda_2 =3$ is: $$E_{\lambda_2}=\bigg\{ \begin{pmatrix}-\frac{4}{3} \\ 1\end{pmatrix}\bigg \}$$