# Determine the principal strain of a 2x2 matrix

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)?

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,$$\begin{bmatrix}1/\sqrt2\\1/\sqrt2\end{bmatrix}=\frac1{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix}$$.
• Thank you for your answer @Shubham Johri. Now where does the $\frac{1}{\sqrt{1^2+1^2}}$ come from? – ecjb Dec 23 '18 at 14:42
• Note that if $\bf v$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda$, $k\bf v$$(k\ne0)$ is also an eigenvector of $A$ corresponding to the same eigenvalue. This is because $A\mathbf v=\lambda\mathbf v\implies A(k\mathbf v)=\lambda(k\mathbf v)$. – Shubham Johri Dec 23 '18 at 14:46
• Take $x=1$ to get the eigenvector $\mathbf v=(1,1)^T$. The norm of $\mathbf v$ is $\|\mathbf v\|=\sqrt{1^2+1^2}$, so if you divide $\bf v$ by $\|\mathbf v\|=\sqrt2$, you will get the unit vector $\frac1{\sqrt2}\begin{bmatrix}1\\1\end{bmatrix}$ which is also an eigenvector corresponding to the same eigenvalue. – Shubham Johri Dec 23 '18 at 14:49