Science:Math Exam Resources/Courses/MATH307/December 2012/Question 03 (b)

MATH307 December 2012

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[hide] Question 03 (b)
Let $S=\left\{[x_{1},x_{2},x_{3}]^{T}:\,x_{1}+x_{2}+x_{3}=0\right\}$ be the subspace of vectors in $\mathbb {R} ^{3}$ whose components sum to zero. (b) Write down a basis for S.

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.

[show] Hint 1
Use the matrix A from part (a) and calculate its kernel as usual.

[show] Hint 2
For an alternative solution, what is the dimension (number of free variables) of S? Fix the free variable(s) and express the remaining variable(s) in terms of those.

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[show] Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If S is the null space of the matrix A from part (a), then the basis for S can be found as the basis of N(A). Let's choose the simplest form $A={\begin{bmatrix}1&1&1\end{bmatrix}}.$ We see that we choose x2 and x3 to be free variables when we set x1 = -x2-x3. That is, the kernel N(A) (and with it the subspace S) can be expressed as ${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}=x_{2}{\begin{bmatrix}-1\\1\\0\end{bmatrix}}+x_{3}{\begin{bmatrix}-1\\0\\1\end{bmatrix}}$ In other words, a basis for S is given by the two vectors $b_{1}={\begin{bmatrix}-1\\1\\0\end{bmatrix}},\quad b_{2}={\begin{bmatrix}-1\\0\\1\end{bmatrix}}$

[show] Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The dimension of S is 2, because if we fix two of x1, x2, x3 we can always choose the remaining such that the equation x1 + x2 + x3 = 0 is satisfied. So let us choose x2 = r and x3 = s. Then x1 = -x2-x3=-r-s. That is, S can be expressed as all vectors of the form ${\begin{bmatrix}-r-s\\r\\s\end{bmatrix}}=r{\begin{bmatrix}-1\\1\\0\end{bmatrix}}+s{\begin{bmatrix}-1\\0\\1\end{bmatrix}}$. In other words, a basis for S is given by the vectors $b_{1}={\begin{bmatrix}-1\\1\\0\end{bmatrix}},\quad b_{2}={\begin{bmatrix}-1\\0\\1\end{bmatrix}}$

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