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Let $\Bbb S=\{\vec x\in\Bbb R^3\mid x_1-x_2=0\}$ and $B=\{(1,2,1),v,(2,1,1)\}$ be a basis of $\Bbb R^3$. It is known that $[(-3,9,1)]_B=\Bigl(\begin{smallmatrix}2\\3\\-1\end{smallmatrix}\Bigr)$.

Find coordinates in the base $B$ of the vectors of a basis of $\Bbb S^\perp$.


I have done the following:

First we have to know $v$. Using the fact that $[(-3,9,1)]_B=\Bigl(\begin{smallmatrix}2\\3\\-1\end{smallmatrix}\Bigr)$, then $$2(1,2,1)+3(v_1,v_2,v_3)+(-1)(2,1,1)=(-3,9,1)\implies\begin{cases}2+3v_1-2=-3,\\4+3v_2-1=9,\\2+3v_3-1=1\end{cases}\equiv\begin{cases}v_1=-1,\\v_2=2,\\v_3=0\end{cases}\implies v=(-1,2,0).$$ To find the orthogonal complement of $\Bbb S$, we need a basis of $\Bbb S$. Let $U_{\Bbb S}$ be that basis. So from $\{\vec x\in\Bbb R^3\mid x_1-x_2=0\}$ we can conclude that $U_{\Bbb S}=\{(1,1,0),(0,0,1)\}$, so to find the coordinates of the basis of $U_{\Bbb S^\perp}$ we have to make two scalar products: $$\begin{cases}(x,y,z)\cdot(1,1,0)=0\\(x,y,z)\cdot(0,0,1)=0\end{cases}\equiv\begin{cases}y=-x,\\z=0,\end{cases}$$ thus $U_{\Bbb S^\perp}=\{(1,-1,0)\}$.

However, this basis IS NOT in the $B$ basis, so we have to state $[(1,-1,0)]_B=\Bigl(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\Bigr)$. Therefore, $$x(1,2,1)+y(-1,2,0)+z(2,1,1)=(1,-1,0)\implies\begin{cases}x-y+2z=1,\\2x+2y+z=-1,\\x+z=0\end{cases}\equiv\begin{cases}x=-1,\\y=0,\\z=1.\end{cases}$$ Hence, the answer to the question is that the coordinates in basis $B$ of a basis of $\Bbb S^\perp$ (we called it $U_{\Bbb S^\perp}$) is $\boxed{(x,y,z)=(-1,0,1)}$.

Are both the reasoning and the words used correct?

Thanks!

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Yes, both reasoning and words used are correct.

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