Science:Math Exam Resources/Courses/MATH221/December 2011/Question 06 (a)

MATH221 December 2011

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Question 06 (a)
Consider the vectors $v_{1}=\left({\begin{array}{c}1/2\\1/2\\1/2\\1/2\end{array}}\right),\quad v_{2}=\left({\begin{array}{c}1/2\\1/2\\-1/2\\-1/2\end{array}}\right),$ $v_{3}=\left({\begin{array}{c}1/2\\-1/2\\1/2\\-1/2\end{array}}\right),\quad v_{4}=\left({\begin{array}{c}1/2\\-1/2\\-1/2\\1/2\end{array}}\right)$ a) Check that v1, v2, v3, v4 is an orthogonal basis of R4.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Check whether the vectors are pairwise orthogonal.

Hint 2
The four vectors will form a basis of $\mathbb {R} ^{4}$ if they are linearly independent.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First we must determine if the vectors are orthogonal. We know that two vectors are orthogonal if their dot product is zero. Consider $v_{1}\cdot v_{2}$. This gives: $(1/2)(1/2)+(1/2)(1/2)+(1/2)(-1/2)+(1/2)(-1/2)=0$ All other dot products are calculated the same way and all vanish. Hence, $v_{1},v_{2},v_{3},v_{4}$ is an orthogonal set. Second, we must determine if the vectors are linearly independent. We will do so by placing them in a matrix and using Gaussian elimination. $\left[{\begin{array}{cccc}1/2&1/2&1/2&1/2\\1/2&1/2&-1/2&-1/2\\1/2&-1/2&1/2&-1/2\\1/2&-1/2&-1/2&1/2\end{array}}\right]$ We multiply the whole matrix by 2 to get $\left[{\begin{array}{cccc}1&1&1&1\\1&1&-1&-1\\1&-1&1&-1\\1&-1&-1&1\end{array}}\right]$ Subtracting the first row from the following three rows, we get $\left[{\begin{array}{cccc}1&1&1&1\\0&0&-2&-2\\0&-2&0&-2\\0&-2&-2&0\end{array}}\right]$ We rearrange the rows so that the second row becomes the last row; each of the last three rows is also multiplied by -1/2. $\left[{\begin{array}{cccc}1&1&1&1\\0&1&0&1\\0&1&1&0\\0&0&1&1\end{array}}\right]$ We subtract row 2 from row 1 and row 2 from row 3 to get $\left[{\begin{array}{cccc}1&0&1&0\\0&1&0&1\\0&0&1&-1\\0&0&1&1\end{array}}\right]$ We subtract row 3 from row 1 and row 3 from row 4 to get $\left[{\begin{array}{cccc}1&0&0&1\\0&1&0&1\\0&0&1&-1\\0&0&0&-2\end{array}}\right]$ Multiplying the last row by -1/2 gives $\left[{\begin{array}{cccc}1&0&0&1\\0&1&0&1\\0&0&1&-1\\0&0&0&1\end{array}}\right]$ Which finally simplifies to $\left[{\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}}\right]$ Because the matrix reduces to an identity matrix the columns must be linearly independent. Thus $v_{1},v_{2},v_{3},v_{4}$ are a basis for $\mathbb {R} ^{4}$.