Find a basis for the eigenspaces corresponding to the eigenvalues
I need help finding an eigenspace corresponding to each eigenvalue of A = $\begin{bmatrix} 1 & -1 & 0 \\ 2 & 4 & 0 \\ 9 & 5 & 4 \end{bmatrix}$ ?
I followed standard eigen-value finding procedures, and I was able to find that $\lambda = 4, 2, 3$. I was even able to find the basis corresponding to $\lambda = 4$:
$\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$
However, I am unable to find the basis corresponding to $\lambda = 2, 3$. I would really appreciate it if someone could please help me with this.
$\lambda = 2$:
First we compute A - I$\lambda$ to get:
$\begin{bmatrix} -1 & -1 & 0 \\ 2 & 2 & 0 \\ 9 & 5 & 2 \end{bmatrix}$
$\endgroup$ 43 Answers
$\begingroup$$\begin{bmatrix} -1 & -1 & 0 \\ 2 & 2 & 0 \\ 9 & 5 & 2 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix}=\mathbf 0$
$x_1 + x_2 = 0\\ x_2 = -x_1\\ 9x_1 + 5(-x_1) + 2x_3 = 0\\ x_3 = -2x_1 $
$\mathbf x = \begin{bmatrix} 1\\ -1\\ -2 \end{bmatrix}$
$\endgroup$ $\begingroup$Remember that the eigenspace of an eigenvalue $\lambda$ is the vector space generated by the corresponding eigenvector.
So, all you need to do is compute the eigenvectors and check how many linearly independent elements you can form from calculating the eigenvector.
$\endgroup$ $\begingroup$You have computed $A - 2 I$. If you cannot see what the nullspace is immediately, you can do row operations to get it into RREF form.
\begin{bmatrix} 1 & 1 & 0\\ 2 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}
Then it is easier to see that the nullspace consists of vectors of the form $(x, -x, -2x)$.
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