Finding Eigenspaces of a Matrix

[Dustinsfl]

$$\begin{bmatrix} 3 & 2\\ 4 & 1 \end{bmatrix}$$
$$det(A-\lambda I)=\begin{vmatrix} 3-\lambda & 2\\ 4 & 1-\lambda \end{vmatrix}=(3-\lambda)(1-\lambda)-8=\lambda^2-4\lambda-5$$
$$\lambda_{1}=5$$ and $$\lambda_{2}=-1$$
When $$\lambda=5$$, $$\begin{bmatrix} -2 & 2\\ 4 & -4 \end{bmatrix}\Rightarrow \begin{bmatrix} 1 & -1\\ 0 & 0 \end{bmatrix}$$
The eigenspace for $$\lambda_{1}$$ is $$\begin{bmatrix} 1\\ 1 \end{bmatrix}$$
When $$\lambda=-1$$, $$\begin{bmatrix} 4 & 2\\ 4 & 2 \end{bmatrix}\Rightarrow \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$
The eigenspace for$$\lambda_{2}$$ is $$\begin{bmatrix} 0\\ 0 \end{bmatrix}$$

I don't know what is going wrong but my second Eigenspace is wrong compared to the books answer which is $$\begin{bmatrix} 1\\ -2 \end{bmatrix}$$

 

[Mark44]

$$\begin{bmatrix} 3 & 2\\ 4 & 1 \end{bmatrix}$$
$$det(A-\lambda I)=\begin{vmatrix} 3-\lambda & 2\\ 4 & 1-\lambda \end{vmatrix}=(3-\lambda)(1-\lambda)-8=\lambda^2-4\lambda-5$$
$$\lambda_{1}=5$$ and $$\lambda_{2}=-1$$
When $$\lambda=5$$, $$\begin{bmatrix} -2 & 2\\ 4 & -4 \end{bmatrix}\Rightarrow \begin{bmatrix} 1 & -1\\ 0 & 0 \end{bmatrix}$$
The eigenspace for $$\lambda_{1}$$ is $$\begin{bmatrix} 1\\ 1 \end{bmatrix}$$
When $$\lambda=-1$$, $$\begin{bmatrix} 4 & 2\\ 4 & 2 \end{bmatrix}\Rightarrow \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$

Your mistake is above. The [4 2; 4 2] matrix doesn't row reduce to the identity matrix. Try again.
Each matrix for calculating the eigenspace can't reduce to the identity; otherwise its determinant would not be zero.

The eigenspace for$$\lambda_{2}$$ is $$\begin{bmatrix} 0\\ 0 \end{bmatrix}$$

I don't know what is going wrong but my second Eigenspace is wrong compared to the books answer which is $$\begin{bmatrix} 1\\ -2 \end{bmatrix}$$

 

[Dustinsfl]

I had a -2 entered into my calc.