# The intersection of $x_1 + x_2 + x_3 + x_4 = 0$ and $x_1 + 2x_2 + 3x_3 + 4x_4 = 0$

Let $\text{U}$ be the intersection of $x_1 + x_2 + x_3 + x_4 = 0$ and $x_1 + 2x_2 + 3x_3 + 4x_4 = 0$ in $\mathbb{R}^4$. Write down the basis $(v_1, v_2)$ of $\text{U}$.

I am confused about how there can exist two vectors that can form a basis as $\text{U}$ only contains the zero vector. Can anyone explain whether the question is wrongly phrased or my logic is incorrect?

• Why do you think it contains only the zero vector? With four dimensions and two linear equations, you still have two degrees of freedom. – Macavity Apr 24 '16 at 9:30
• I thought that it contains only the zero vector since the only place where both these equations are equal are when $x_1=x_2=x_3=x_4=0$. – Dheeraj Putta Apr 24 '16 at 9:35
• @DheerajPutta No. Write down the 2 row, 4 column matrix and bring in standard form. There are many other solutions. – Henno Brandsma Apr 24 '16 at 9:36
• How would i write the matrix $$\begin{pmatrix}x_1 & x_2 & x_3 & x_4 \\ x_1 & 2x_2 & 3x_3 & 4x_4 \end{pmatrix}$$ in standard form? – Dheeraj Putta Apr 24 '16 at 9:49
• Hint: Any linear combination of $(1, -1, -1, 1)$ and $(1, 0, -3, 2)$ for eg satisfi s both. – Macavity Apr 24 '16 at 9:55

Set $x_3 = a, x_4 = b$ as free parameters, then $x_1 + x_2 + a + b = 0$, $x_1 + 2x_2 + 3a + 4b = 0$.
Substracting the first from the second: $x_2 + 2a + 3b = 0$, so $x_2 = -2a - 3b$. We substitute this in the first one to get $x_1 = -x_2 - a - b = 2a + 3b - a -b = a + 2b$.
So all vectors of the form $(a+2b, -2a-3b, a, b)$ are solutions, which can be written as $a(1,-2,1,0) + b(2, -3, 0, 1)$, which shows us a base...