find a basis of r3 containing the vectors

\[\left\{ \left[ \begin{array}{r} 1 \\ 2 \\ -1 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} 1 \\ 3 \\ -1 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 1 \\ 3 \\ 0 \\ 1 \end{array} \right] \right\}\nonumber \] Since the first, second, and fifth columns are obviously a basis for the column space of the , the same is true for the matrix having the given vectors as columns. I would like for someone to verify my logic for solving this and help me develop a proof. If it has rows that are independent, or span the set of all \(1 \times n\) vectors, then \(A\) is invertible. The column space is the span of the first three columns in the original matrix, \[\mathrm{col}(A) = \mathrm{span} \left\{ \left[ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array} \right], \; \left[ \begin{array}{r} 2 \\ 3 \\ 2 \\ 3 \end{array} \right] , \; \left[ \begin{array}{r} 1 \\ 6 \\ 1 \\ 2 \end{array} \right] \right\}\nonumber \]. Let \(\dim(V) = r\). The tools of spanning, linear independence and basis are exactly what is needed to answer these and similar questions and are the focus of this section. Find a basis for $A^\bot = null (A)^T$: Digression: I have memorized that when looking for a basis of $A^\bot$, we put the orthogonal vectors as the rows of a matrix, but I do not know why we put them as the rows and not the columns. 0But sometimes it can be more subtle. Describe the span of the vectors \(\vec{u}=\left[ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right]^T\) and \(\vec{v}=\left[ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right]^T \in \mathbb{R}^{3}\). Then it follows that \(V\) is a subset of \(W\). Notice that the vector equation is . 2 of vectors (x,y,z) R3 such that x+y z = 0 and 2y 3z = 0. Consider the vectors \(\vec{u}, \vec{v}\), and \(\vec{w}\) discussed above. Why do we kill some animals but not others? Otherwise, pick \(\vec{u}_{3}\) not in \(\mathrm{span}\left\{ \vec{u}_{1},\vec{u}_{2}\right\} .\) Continue this way. Who are the experts? Question: 1. It turns out that this is not a coincidence, and this essential result is referred to as the Rank Theorem and is given now. Learn how your comment data is processed. If \(V= \mathrm{span}\left\{ \vec{u}_{1}\right\} ,\) then you have found your list of vectors and are done. Find Orthogonal Basis / Find Value of Linear Transformation, Describe the Range of the Matrix Using the Definition of the Range, The Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero, Condition that Two Matrices are Row Equivalent, Quiz 9. Therefore the nullity of \(A\) is \(1\). Let \(V\) be a nonempty collection of vectors in \(\mathbb{R}^{n}.\) Then \(V\) is a subspace of \(\mathbb{R}^{n}\) if and only if there exist vectors \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\) in \(V\) such that \[V= \mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}\nonumber \] Furthermore, let \(W\) be another subspace of \(\mathbb{R}^n\) and suppose \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\} \in W\). Find a basis for each of these subspaces of R4. Then \(A\) has rank \(r \leq n

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