# Linear algebra I

### Basic information

You can always reach me by e-mail: 1@2354knopkammffczcuni.

### Credit for the class will be awarded for at least $2/3$ of all possible points for homeworks and tests (each 10 points)

Actual points (Winter semester) Actual points (Summer semester)### Other resources

- MIT OpenCourseWare Linear algebra by Prof. Gilbert Strang main site
- The sketch proof of the fact, that two vectors $\bu = (u_1, u_2)$ and $\bv = (v_1, v_2)$ in $\R^2$ (i.e. in the plane) are perpendicular if either $v_1 = u_2, v_2 = -u_1$ or $v_1 = -u_2, v_2 = u_1$.
- Use the fact, that $cos\varphi = \frac{\bu\cdot\bv}{\|\bu\|\|\bv\|}$, where $\varphi$ is the angle between vectors $\bu$ and $\bv$.
- We want $\bu$ and $\bv$ to be perpendicular, so $cos\varphi = 0$, moreover we would like to have vectors of the same length.
- This drives us to the form $\bu\cdot\bv = \|\bu\|^2$.
- From the definition of the dot product ($\cdot$) and Eukliedian norm we get $u_1v_1 + u_2v_2 = 0$.
- This gives us $u_1v_1 = -u_2v_2$. Which is the same as $\frac{v_1}{u_2} = -\frac{v_2}{u_1}$. Now giving the left-hand side value $1$ and $-1$ gives us the result.