Show that the column space of a matrix is not equal to $\mathbb R^3$
Solution 1
By the rank-nullity theorem you know that
$$rank(A) + null(A) = n$$
Find the rank of $A$. (the number of linearly independent columns in $A$) to get the dimension of the column space of $A$.
Hint
How many vectors do you need to span $\mathbb{R}^3$? (What is the dimension of $\mathbb{R}^3$?)
What does the rank of $A$ tell you then? And what does it need to be for the columnspace of $A$ to span $\mathbb{R}^3$?
Solution 2
Simple. If it is equal to $\mathbb{R}^3$ then $\exists$ $\alpha,\beta,\gamma\in\mathbb{R}$ such that $$ \alpha\left(\begin{array}{c} 4\\ 0\\ 5 \end{array}\right)+\beta\left(\begin{array}{c} -1\\ 0\\ -1 \end{array}\right)+\gamma\left(\begin{array}{c} 2\\ 0\\ 6 \end{array}\right) = \left(\begin{array}{c} a\\ b\\ c \end{array}\right) $$ for all $a,b,c\in\mathbb{R}$. But $$ \alpha\left(\begin{array}{c} 4\\ 0\\ 5 \end{array}\right)+\beta\left(\begin{array}{c} -1\\ 0\\ -1 \end{array}\right)+\gamma\left(\begin{array}{c} 2\\ 0\\ 6 \end{array}\right) \neq \left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right) $$ for any $\alpha,\beta,\gamma\in\mathbb{R}$.
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Alex Mathers
Updated on April 15, 2020Comments
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Alex Mathers over 3 years
Show that the column space of \begin{pmatrix} 4 &−1& 2 \\ 0 &0& 0 \\ 5 &−1 &6 \\ \end{pmatrix} is not equal to $\mathbb R^3$.
I have begun by setting my vector $(x_1,x_2,x_3)=(0,0,0)$, giving $$4x_1-x_2+2x_3=0 \\ 5x_1-x_2+6x_3=0 $$ and attempted to row reduce using rref. This resulted in $x_1-4x_3,x_2=-14x_3$. I am not sure if this answers the question or if I have even gone about it the right way.
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peterwhy over 7 yearsDo you think $\begin{pmatrix} 0\\1\\0 \end{pmatrix}$ is in the column space?
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Aritra Das over 7 yearsI'm note entirely sure, but could this also be done using row rank = column rank ? (Since there is a zero row, row rank is less than 3 and hence column rank is less than 3 and thus column space must have dimension less than 3 whereas $\mathbb{R}^3$ has dimension 3)
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David Simmons over 7 yearsCorrect. The simplest way is to just take the determinant of the matrix. If this is non zero, the columns are linearly independent and thus span the space. Otherwise, they do not span the space.
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David Simmons over 7 yearsI like to think of the determinant as a functional that `determines' whether vectors are linearly dependent. If the are, the determinant is zero.
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Aritra Das over 7 yearsAt last, now I know why determinant is called determinant! Thank you very much for your replies. I've currently exceeded my daily vote limit and hence couldn't upvote your answer or comments. I'll be sure to do that tomorrow. Hope you don't mind.