The matrix \(A =\) \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) has the eigenvector and eigenvalues,
Eigenvalue with eigenvector
Eigenvalue with eigenvector
Please do recall that eigenvectors are not unique. As long as your answer is proportional to the answer provided its okay. For example, if \(\textbf{x}\) is an eigenvector with eigenvalue \(\lambda\), then for any scalar \(c \not= 0 \), \(c \textbf{x}\) is still an eigenvector with eigenvalue \(\lambda\). If you got a different answer \(\textbf{y}\), you can check that there is a \(c \not= 0 \) so that \(\textbf{x} = c \textbf{y}\). If there is, then your answer is good! Of course the best way to check your work is to directly compute \(A \textbf{x}\) and see if it really is an eigenvector with the correct eigenvalue.
Okay, first thing you want to do is find the eigenvalues by finding roots of \( p(\lambda ) = \det ( A - \lambda I ) =0 \).
For us this gives, $$ \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
This gives us the eigenvalues, and .
Now we just need to find the eigenvector associated to each eigenvalue.
Lets start with
Set \( \textbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \) and compute,
The two equations are redundant. We can see they are solved for \(v_1 = \) and \(v_2 = \) which gives .
Now we move onto the second eigenvalue,
Set \( \textbf{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \) and compute,
Again, the two equations are redundant. We can see they are solved for \(u_1 = \) and \(u_2 = \) which gives .