47edb6…: “As opposed to the spring disk”

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True/False: If $\hat {x}$ is a least-squares solution of $A\vec {x} = \vec {b}$ , then $A\hat {x}$ is a point in the column space of $A$ closest to $\vec {b}$ .

True False

When does $A\vec {x} = \vec {b}$ have exactly one least-squares solution? Select all that apply.

When $A$ is $n\times n$ . When $A\vec {x} = \vec {b}$ is consistent. When the columns of $A$ are linearly independent. When $A\vec {x} = \vec {b}$ is has exactly one solution.

Let $A = \begin {bmatrix} 4 &1\\0&2\\0&1\end {bmatrix}$ and $\vec {b}= \begin {bmatrix}1\\-1\\2\end {bmatrix}$ .

Compute $A^TA$ . $A^TA =\begin {bmatrix} \answer {16}& \answer {4}\\\answer {4}& \answer {6}\end {bmatrix}$
Compute $A^T\vec {b}$ . $A^T\vec {b} = \begin {bmatrix} \answer {4}\\\answer {1}\end {bmatrix}$
Solve the normal equations $A^TA\vec {x} = A^T\vec {b}$ to determine the set of least-squares solutions.
$\vec {x} = \begin {bmatrix}1/4\\ \answer {0}\end {bmatrix}$

Why is there only one least-squares solution?

Because the columns of $A$ are linearly independent. Because $A\vec {x} = \vec {b}$ is has exactly one solution.

Let $A = \begin {bmatrix} 1 &-1&2\\0&2&0\\0&5&0\end {bmatrix}$ and $\vec {b}= \begin {bmatrix}3\\4\\1\end {bmatrix}$ .

Compute $A^TA$ . $A^TA =\begin {bmatrix} \answer {1}& \answer {-1}&\answer {2}\\\answer {-1}& \answer {30}&\answer {-2}\\\answer {2}&\answer {-2}&\answer {4}\end {bmatrix}$
Compute $A^T\vec {b}$ . $A^T\vec {b} = \begin {bmatrix} \answer {3}\\\answer {10}\\\answer {6}\end {bmatrix}$
Solve the normal equations $A^TA\vec {x} = A^T\vec {b}$ to determine the set of least-squares solutions.
$\vec {x} = x_3\begin {bmatrix}\answer {-2}\\ \answer {0}\\\answer {1}\end {bmatrix}+ \begin {bmatrix}\answer {100}/29\\ \answer {13}/29\\\answer {0}\end {bmatrix}$

When will the method of solving the normal equations, $A^TA\vec {x} = A^T\vec {b}$ , work for finding the least-squares solutions?

Always Only when the columns of $A$ are linearly independent. Only when the columns of $A$ are orthogonal.

When will the method of finding the orthogonal projection of $\vec {b}$ work for finding the least-squares solutions?

Always Only when the columns of $A$ are linearly independent. Only when the columns of $A$ are orthogonal.

When will the method of usng the $QR$ factorization work for finding the least-squares solutions?

Always Only when the columns of $A$ are linearly independent. Only when the columns of $A$ are orthogonal.