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I am attempting to solve the following problem.

I am given the matrix A=\begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 \\ \end{pmatrix}

Specifically I am asked to: (1) find the dimension of the range of this matrix, (2) give a range for this subspace, (3) find the dimension of the null space of this matrix, and (4) give a basis for this subspace.

I just keep staring at the problem but can't seem to put my finger on how to tackle it since I am not given a linear transformation.

Thanks in advance for the help!

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  • $\begingroup$ Just regard $A$ as a linear transformation. Then, range will be a set $\{A\mathbf{x}:\mathbf{x}\in\mathbb{R}^4\}$ and null space be $\{\mathbf{x}\in\mathbb{R}^4:A\mathbf{x}=\mathbf{0}\}$. $\endgroup$ Feb 19 '16 at 11:07
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Note that the $\ker A$ has dimension $3$ and is spanned by vector $(1,0,0,-1) ,(0,1,0,-1) , (0,0,1,-1)$ while $ Im A$ is spanned by vector $(1,1,1,1)$ Indeed to find $\ker A$ you have to solve a linear system $$\begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 \\ \end{pmatrix}\cdot \begin{pmatrix} X_1 \\ X_2 \\ X_3 \\ X_4\\ \end{pmatrix}= vector zero$$ Therefore you find that $X_1+X_2+X_3+X_4=0$

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  • $\begingroup$ So when you mention that we have to solve a linear system to find ker A, are you saying that is how you find those spanning vectors? (1,0,0-1), (0,1,0,-1), and (0,0,1,-1)? Also, since you only have one linear equation to work with, are you basically just playing around with numbers and seeing what works? $\endgroup$
    – King Tut
    Feb 19 '16 at 16:50
  • $\begingroup$ @KingTut No, you don’t “just play around with numbers,” although in this case it is pretty easy so come up with a solution by inspection. You can use any of several methods to solve this system of linear equations, such as row-reduction. Once you have the row-reduced echelon form of the matrix, you can basically read off a basis for its kernel. $\endgroup$
    – amd
    Feb 19 '16 at 18:44
  • $\begingroup$ @amd Could you please provide such an example? $\endgroup$
    – King Tut
    Feb 19 '16 at 19:00
  • $\begingroup$ @KingTut There are several examples here. $\endgroup$
    – amd
    Feb 19 '16 at 21:33
  • $\begingroup$ @amd Thanks amd! $\endgroup$
    – King Tut
    Feb 20 '16 at 21:01

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