Show that the column space of a matrix is not equal to $\mathbb R^3$

1,948

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 $\alpha,\beta,\gamma\in\mathbb{R}$.

Share:
1,948

Related videos on Youtube

Alex Mathers
Author by

Alex Mathers

Updated on April 15, 2020

Comments

  • Alex Mathers
    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.

    • peterwhy
      peterwhy over 7 years
      Do you think $\begin{pmatrix} 0\\1\\0 \end{pmatrix}$ is in the column space?
  • Aritra Das
    Aritra Das over 7 years
    I'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)
  • David Simmons
    David Simmons over 7 years
    Correct. 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.
  • David Simmons
    David Simmons over 7 years
    I like to think of the determinant as a functional that `determines' whether vectors are linearly dependent. If the are, the determinant is zero.
  • Aritra Das
    Aritra Das over 7 years
    At 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.