Understanding Eigenvectors: Solving for Eigenvectors of a 2x2 Matrix

[mprm86]

The eigenvalues of the matrix $$\left(\begin{array}{cc}0 & \frac{1}{2}\\\frac{1}{2} & 0\end{array}\right)$$ are $\lambda_1 = \frac{1}{2}$ and $\lambda_2 = -\frac{1}{2}$
The problem here is that I have no idea of how to calculate the eigenvectors. Could some one please explain me, in detail, how do I find the eigenvectors?
Thanks in advance.

 

[Hurkyl]

To find the eigenvectors of the matrix A corresponding to the eigenvalue λ, you simply solve the equation Av = λv for v.

 

[tacman]

Sure, I'd be happy to explain how to find the eigenvectors for this matrix. First, let's review what eigenvectors are.

Eigenvectors are special vectors that, when multiplied by a square matrix, result in a scalar multiple of that same vector. In other words, they are vectors that do not change direction when multiplied by a matrix, but only get scaled by a certain factor (known as the eigenvalue).

To find the eigenvectors of a 2x2 matrix, we can follow a simple process:

1. Start by subtracting the eigenvalue from the main diagonal of the matrix. In this case, we would subtract λ from the top left and bottom right elements of the matrix. So, we would get:

\left(\begin{array}{cc}-\lambda & \frac{1}{2}\\\frac{1}{2} & -\lambda\end{array}\right)

2. Next, we need to find the determinant of this new matrix. The determinant of a 2x2 matrix is calculated by multiplying the top left and bottom right elements, then subtracting the product of the top right and bottom left elements. So, in this case, the determinant would be:

det = (-\lambda)(-\lambda) - (\frac{1}{2})(\frac{1}{2}) = \lambda^2 - \frac{1}{4}

3. Now, we need to set the determinant equal to 0 and solve for λ. In this case, we would have:

\lambda^2 - \frac{1}{4} = 0

Solving for λ, we get two solutions: λ = \frac{1}{2} and λ = -\frac{1}{2}.

4. Now that we have our eigenvalues, we can find the corresponding eigenvectors. To do this, we plug each eigenvalue back into the original matrix (not the one we subtracted from) and solve for x and y in the equation Ax = \lambda x. This will give us two equations, one for each eigenvalue. In this case, we would have:

For λ = \frac{1}{2}:

\left(\begin{array}{cc}0 & \frac{1}{2}\\\frac{1}{2} & 0\end{array}\right) \left(\begin{array}{c}