Question

Consider the subspace $\mathit{L}$of ${\mathbf{ℝ}}^{\mathbf{5}}$ spanned by the given vector. Find a basis of ${\mathit{L}}^{\mathbf{\perp }}$ . See Exercise 53.

$\left[\begin{array}{c}1\\ 2\\ 3\\ 4\\ 5\end{array}\right]$.

Short answer

The basis for $L^{\perp }$is,$\left\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-5\\ 0\\ 0\\ 0\\ 1\end{array}\right]\right\}$

Step-by-step solution

Consider the set

The vectors in ${\mathrm{ℝ}}^n$are linearly dependent if and only if there are nontrivial relation among them.

If linearly dependent, then, the vectors are said to be redundant.

Consider the vector$\left[\begin{array}{c}1\\ 2\\ 3\\ 4\\ 5\end{array}\right]$

Consider the condition

Let $\overrightarrow{w}·\overrightarrow{v}=0,\overrightarrow{w}\in L^{\perp },\overrightarrow{v}\in V$

Then, $L^{\perp }$must be orthogonal to $L$

If, $L=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\\ 5\end{array}\right]$then, $L^{\perp }=\left[1 2 3 4 5\right]$

$$\begin{gathered} \left[1 2 3 4 5\right]·\left[\begin{array}{c}1\\ 2\\ 3\\ 4\\ 5\end{array}\right]=0 \\ {x}_1+2x_2+3x_3+4x_4+5x_5=0 \end{gathered}$$

Compute the values

If, $x_3=x_4=x_5=0$

Then, $x_1=-2x_2$.

Thus, $x_1=-2,x_2=1,x_3,x_4,x_5=0$.

localid="1664208677638" $x=\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right]$

If, $x_2=x_4=x_5=0$

Then, $x_1=-3x_3$

Thus, $x_1=-3,x_3=1,x_2,x_4,x_5=0$

localid="1664208625829" $x=\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\\ 0\end{array}\right]$

If, $x_2=x_3=x_5=0$

Then, $x_1=-4x_4$

Thus,$$\begin{gathered} x=\left[-4 \\ 0 \\ 0 \\ 1 \\ 0\right] \end{gathered}$$$x_1=-4,x_4=1,x_2,x_3,x_5=0$

If, $x_2=x_3=x_4=0$

Then, $x_1=-5x_5$

Thus, $x_1=-5,x_5=1,x_2,x_3,x_4=0$

localid="1664208654026" $x=\left[\begin{array}{c}-5\\ 0\\ 0\\ 0\\ 1\end{array}\right]$

Final answer

The basis for $L^{\perp }$is, role="math" localid="1664208383236" $\left\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-5\\ 0\\ 0\\ 0\\ 1\end{array}\right]\right\}$.