Science:Math Exam Resources/Courses/MATH307/April 2006/Question 05 (b)

MATH307 April 2006

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[hide] Question 05 (b)
A matrix A and a vector b are given by $\displaystyle A={\begin{bmatrix}1&0\\1&1\\0&1\end{bmatrix}}$ $\displaystyle b={\begin{bmatrix}2\\3\\4\end{bmatrix}}$ Find an orthonormal basis for $\mathbb {R} ^{3}:\{u_{1},u_{2},u_{3}\}$ in which $u_{1},u_{2}$ span the column space R(A) of the matrix A. How does the third basis vector $u_{3}$ relate to the fundamental subspaces of A?

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[show] Hint

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[show] Solution

$u_{3}$ is orthogonal to the R(A) since it is a part of the orthonormal basis for $\mathbb {R} ^{3}$ Recall that $R(A)^{\perp }=N(A^{T})$ Since the dim of $N(A^{T})=1$ , the basis of $N(A^{T})$ is $u_{3}$ We can find $u_{3}$ either by finding $N(A^{T})$ or using gram-schmidt process or with cross products We solve it by finding $N(A^{T})$ $A^{T}u_{3}=0$ ${\begin{bmatrix}1&1&0\\0&1&1\end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$ $a=c$ $b=-a$ $N(A^{T})=\{{\begin{bmatrix}1\\-1\\1\end{bmatrix}}\}$ Normalize the vector $u_{3}={\frac {1}{\sqrt {3}}}{\begin{bmatrix}1\\-1\\1\end{bmatrix}}$ Alternatively, if you did not recognize that $u_{3}\in N(A^{T})$ we can still compute $u_{3}$ with cross product $u_{3}=u_{1}\times u_{2}={\begin{vmatrix}i&j&k\\{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}&0\\-{\frac {\sqrt {2}}{2{\sqrt {3}}}}&{\frac {\sqrt {2}}{2{\sqrt {3}}}}&{\sqrt {\frac {2}{3}}}\end{vmatrix}}$ $={\frac {1}{\sqrt {3}}}{\begin{bmatrix}-1\\1\\-1\end{bmatrix}}$ And we note our answer from computing the nullspace versus the cross product is a multiple of -1 but either one can be used as $u_{3}$ for the orthonormal basis of $\mathbb {R} ^{3}$