Science:Math Exam Resources/Courses/MATH307/April 2013/Question 04 (a)

MATH307 April 2013

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Question 04 (a)
Let S be the subspace of $\mathbb {R} ^{3}$ spanned by ${\begin{bmatrix}1\\1\\1\end{bmatrix}}$, ${\begin{bmatrix}2\\-1\\-1\end{bmatrix}}$ and ${\begin{bmatrix}4\\1\\1\end{bmatrix}}$. Given the MATLAB/Octave calculation > rref([1 2 4; 1 -1 1; 1 -1 -1]) ans = 1 0 2 0 1 1 0 0 0 (a) Find the matrix P that projects onto S.

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Hint

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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. We are given ${\begin{bmatrix}1&2&4\\1&-1&1\\1&-1&1\end{bmatrix}}$ and its reduced row echelon form ${\begin{bmatrix}1&0&2\\0&1&1\\0&0&0\end{bmatrix}}$ If we take the original matrix ${\begin{bmatrix}1&2&4\\1&-1&1\\1&-1&1\end{bmatrix}}$ as A, then $A^{T}A$ is not invertible, so we cannot obtain P. So, we need to find another matrix. Since column 1 and column 2 are pivot columns, we take corresponding columns of A. Those are ${\begin{bmatrix}1\\1\\1\end{bmatrix}}$ and ${\begin{bmatrix}2\\-1\\-1\end{bmatrix}}$, and they are linearly independent. So, we define a new matrix $A={\begin{bmatrix}1&2\\1&-1\\1&-1\end{bmatrix}}$ $A^{T}={\begin{bmatrix}1&1&1\\2&-1&-1\end{bmatrix}}$ Projection matrix is given by $P=A(A^{T}A)^{-1}A^{T}$ So, if we substitute $A$ and $A^{T}$, we get $P=A(A^{T}A)^{-1}A^{T}={\begin{bmatrix}1&2\\1&-1\\1&-1\end{bmatrix}}\left({\begin{bmatrix}1&1&1\\2&-1&-1\end{bmatrix}}{\begin{bmatrix}1&2\\1&-1\\1&-1\end{bmatrix}}\right)^{-1}{\begin{bmatrix}1&1&1\\2&-1&-1\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&{\frac {1}{2}}&{\frac {1}{2}}\\0&{\frac {1}{2}}&{\frac {1}{2}}\end{bmatrix}}$