# Explore Linear Transformations and Eigenvectors

In the graph below, you can control the output vectors for $$\mathbf e_1$$ (red) and for $$\mathbf e_2$$ (blue). These two vectors completely determine the linear transformation. You can then control the input vector $$\mathbf x$$ (purple), and the output vector $$T(\mathbf x)$$ is then updated. The ratio of lengths $$|T(\mathbf x)|/|\mathbf x|$$ and the angle between $$T(\mathbf x)$$ and the span of $$\mathbf x$$ are also computed.

$$A = \begin{bmatrix} 1.75 & 0.75 \\ 0.75 & 1.75 \end{bmatrix}$$
$$\mathbf x = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
$$T(\mathbf x) = \begin{bmatrix} 2.5 \\ 2.5 \end{bmatrix}$$
$$\displaystyle \frac{|T(\mathbf x)|}{|\mathbf x|} = 2.5$$
Angle between span of $$\mathbf x$$ and $$T(\mathbf x)$$: 0 degrees."

## Discussion

The vector $$\mathbf x$$ is an eigenvector if $$T(\mathbf x)$$ is in the span of $$\mathbf x$$. This can be seen in the demo above if the angle between $$T(\mathbf x)$$ and the span of $$\mathbf x$$ is zero. When $$\mathbf x$$ is an eigenvector, we must have $$T(\mathbf x) = \lambda \mathbf x$$ for some scalar $$\lambda$$. That scalar is called the eigenvalue associated with the eigenvector $$\mathbf x$$. This can be seen in the demo above when you have found an eigenvector by computing the ratio of the lengths, but with a negative sign if the vectors are pointing in opposite directions.

Here are some example matrices $$A$$ to explore. Move the image vectors $$T(\mathbf e_1)$$ and $$T(\mathbf e_2)$$ to set the matrix (as close as possible). Then move the vector $$\mathbf x$$ to see if you can determine any eigenvectors and the corresponding eigenvalues for the matrices.

1. $$A = \begin{bmatrix} 1.75 & 0.75 \\ 0.75 & 1.75 \end{bmatrix}$$
2. $$A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
3. $$A = \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}$$
4. $$A = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$$
5. $$A = \begin{bmatrix} 1 & 0.5 \\ 2 & 1 \end{bmatrix}$$