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I was given the following question:

The given set is a basis for subset $W$. Use the Gram-Schmidt process to produce an orthogonal basis for $W$.$$\left\{\left[\begin{matrix}0\\8\\8\\\end{matrix}\right],\left[\begin{matrix}4\\7\\3\\\end{matrix}\right]\right\}$$

The Gram-Schmidt process is a method of turning a basis $(x_1,x_2)$ into an orthogonal basis $(v_1,v_2)$. The first vector can remain the same, making $v_1=\left(\begin{matrix}0\\8\\8\\\end{matrix}\right)$. The method for finding the second vector is to take the projection of the $x_2$ on $v_1$ and subtract it from $x_2$. This looks like this: $$\vec{x}_2-\left[\frac{\vec{x}_2\bullet\vec{v}_1}{\vec{v}_1\bullet\vec{v}_1}\right]\vec{v}_1\\\vec{x}_2\cdot\vec{v_1}=0+54+24=80\\\vec{v}_1\cdot\vec{v}_1=0+64+64=128\\\frac{80}{128}\vec{v}_1=\left(\begin{matrix}0\\5\\5\\\end{matrix}\right)\\\vec{x}_2-\left(\begin{matrix}0\\5\\5\\\end{matrix}\right)=\left(\begin{matrix}4\\2\\-2\\\end{matrix}\right)$$My final solution, therefore, is: $\vec{v}_1=\left(\begin{matrix}0\\8\\8\\\end{matrix}\right),\vec{v}_2=\left(\begin{matrix}4\\2\\-2\\\end{matrix}\right)$.

I feel confident about my work (maybe in error :)) but the book has a different result: $$\vec{v}_1=\left(\begin{matrix}0\\8\\8\\\end{matrix}\right),\vec{v}_2=\left(\begin{matrix}4\\5\\-5\\\end{matrix}\right)$$I do see how both my solution and the books solution creates an orthogonal basis, but did I make a mistake somewhere?

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You are right, and the book is wrong.

Indeed the second vector given by the book, $$\left(\begin{matrix}4\\5\\-5\\\end{matrix}\right)$$

does not belong to $W$.

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  • $\begingroup$ This may be a very basic question, but how do you know that (4,5,-5) does not belong to W? $\endgroup$
    – Burt
    Sep 1, 2020 at 20:57
  • $\begingroup$ Try to find $a,b \in \mathbb{R}$ so that vector is equal to $au_1 + bu_2$, where $u_1$ and $u_2$ are the two basis vectors of $W$... $\endgroup$ Sep 1, 2020 at 20:58
  • $\begingroup$ I can set it up in a matrix with 4,5,5 as the solution matrix augmented to the matrix made up of the basis vectors? $\endgroup$
    – Burt
    Sep 1, 2020 at 20:59
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    $\begingroup$ Yes, if you want. Or try to solve the system "by hand" (it is enough simple to deal with it by hand). You should find that there is no solution. $\endgroup$ Sep 1, 2020 at 21:00

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