Determine the principal strain of a 2x2 matrix

Question

For a 2D problem the strain matrix is given by

$$ \begin{bmatrix} \varepsilon_{xx} & \varepsilon_{xy} \\ \varepsilon_{xy} & \varepsilon_{yy} \\ \end{bmatrix} = \begin{bmatrix} 0 & 0.1 \\ 0.1 & 0 \\ \end{bmatrix} $$

Determine the principal strains

Solution:

First the eigenvalues have to be computed:

\begin{align} \det{\textbf{A} - \lambda \mathbb{I}} &= \det \begin{bmatrix} - \lambda & 0.1 \\ 0.1 & -\lambda \end{bmatrix} = \lambda^2 - 0.1^2 = 0\\ \lambda_{1,2} &= \pm 0.1 \end{align}

Then the first eignevector $\vec{v}_1$ is determined by

\begin{align} \begin{bmatrix}-0.1 & 0.1 \\ 0.1 & -0.1 \\ \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ -0.1x + 0.1y &= 0 \label{eq1} \tag{1}\\ \vec{v}_1 & = \frac{1}{\sqrt{1^2 + 1^2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \label{eq2} \tag{2} \end{align}

and the second eigenvector is given by

$$ \vec{v}_2 \begin{pmatrix} -1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} $$

My question

This is certainly a basic algebra question, but how to you get from (1) to (2)?

Answer 1

You have $0.1y-0.1x=0\implies x=y$. The eigenvector $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\x\end{bmatrix},x\in\Bbb R-\{0\}$. So one possibility for the eigenvector is when $x=1/\sqrt2$, that is,$\begin{bmatrix}1/\sqrt2\\1/\sqrt2\end{bmatrix}=\frac1{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix}$.