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Let $U$ be the subspace of $\mathbb{C}^{5}$ such that

$$U = \{(z_1,z_2,z_3,z_4,z_5): (6z_1 = z_2)\wedge(z_3+2z_4+3z_5=0)\}$$

Find a basis of $U$.

Here is what I came up with:

Claim: $(1,6,0,0,0),(0,0,-2,1,0),(0,0,-3,0,1)$ is a basis.

Now I can see that the above list of vectors is a basis but I don't know how to start by proving it.

Here is what I have tried:

WTS for all $z\in U$, $z = a_1(1,6,0,0,0) + a_2(0,0,-2,1,0) + a_3(0,0,-3,0,1)$, forall $a_i\in\mathbb{F}$.

So let $z\in U$, then \begin{align*} z & = (z_1,6z_1,z_3,z_4,z_5)\\\\ & = (z_1,6z_1,0,0,0)+(0,0,-2z_4,z_4,0) + (0,0,-3z_5,0,z_5)\\\\ & = z_1(1,6,0,0,0)+z_4(0,0,-2,1,0)+z_5(0,0,-3,0,1) \end{align*}

But this doesn't seem right, unless $\mathbb{F} = \mathbb{C}$.

Átila Correia
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Remu X
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1 Answers1

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$$ U = \{(z_1,z_2,z_3,z_4,z_5): (6z_1 = z_2)\wedge(z_3+2z_4+3z_5=0)\} $$

For any vector $z \in U$, we find that $$ z_2 = 6 z_1 $$ and $$ z_3 = -2 z_4 - 3 z_5 $$

Hence, it follows that $$ z = \left[ \begin{array}{c} z_1 \\ z_2 \\ z_3 \\ z_4 \\ z_5 \\ \end{array} \right] = \left[ \begin{array}{c} z_1 \\ 6 z_1 \\ - 2 z_4 - 3 z_5 \\ z_4 \\ z_5 \\ \end{array} \right] = z_1 \mathbf{v}_1 + z_4 \mathbf{v}_2 + z_5 \mathbf{v}_3 $$ where $$ \mathbf{v}_1 = \left[ \begin{array}{c} 1 \\ 6 \\ 0 \\ 0 \\ 0 \\ \end{array} \right], \ \ \mathbf{v}_2 = \left[ \begin{array}{c} 0 \\ 0 \\ -2 \\ 1 \\ 0 \\ \end{array} \right], \ \ \mathbf{v}_3 = \left[ \begin{array}{c} 0 \\ 0 \\ -3 \\ 0 \\ 1 \\ \end{array} \right] $$

Since $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent, we conclude that $$ \mathcal{B} = \{ \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \} $$ is a basis for $U$. Hence, $\mbox{dim}(U) = 3$.

Sundar
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