# Gaussian Elimination: Find a matrix satisfying the conditions.

QUESTION
Let $\begin{array}{cc} x=[1&2&3&4]^T \end{array}$. Find a matrix $E = I - \alpha e_i e_4^T$

such that $Ex =$ $\begin{bmatrix} 1\\ 2\\ 0\\ 4\end{bmatrix}$

Any help with this would be appreciated. The part I'm most confused about is the $e_i$ and $e_4^T$. What do these represent? I think I should be fine after that I'm just a little confused at the moment.

• what are the dimensions of $e_i$ and $e_4^T$? Do you know what a basis vector is? Commented Mar 25, 2018 at 5:50
• I've been told that $e_4^T$ is the vector [0 0 0 1]^T
– user544141
Commented Mar 25, 2018 at 6:33

Usually, the notation $e_j$, or sometimes $\hat{e}_j$, is used to denote the standard basis vectors of $\mathbb{R}^n$; $e_j$ is a vector that has a $1$ in the $j$-th component and $0$ in the remaining components. E.g.

$$e_3 = \left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix} \right]$$

Do be aware, however, that many sources don't distinguish very well between whether a vector should appear as a row or a column when treated as a matrix. Don't be surprised to see some sources use $e_3$ to mean $\left[ \begin{matrix} 0 & 0 & 1 & 0 \end{matrix} \right]$ instead, or even switch back and forth between the two options!

• Thanks! This was really helpful!
– user544141
Commented Mar 25, 2018 at 23:50

$$Ex=(I-\alpha e_i e^{\text{T}}_4)x =Ix -\alpha e_i (e^{\text{T}}_4 x) =x-\alpha x_4 e_i.$$ $$(Ex)_3 =x_3-\alpha x_4 (e_i)_3.$$ Hint. For what values of $\alpha \in \mathbb{R}$ and $i \in \{1, 2, 3, 4\}$ does this equality $$(Ex)_3 =x_3-\alpha x_4 (e_i)_3 =0$$ hold?

$$\alpha=\frac{x_3}{x_4}=\frac34,\quad i=3.$$