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.

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. \(A = \begin{bmatrix} 1 & 0.5 \\ 2 & 1 \end{bmatrix}\)