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Let \begin{eqnarray} \begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1 \end{pmatrix} \end{eqnarray} be the matrix of a linear map $T$. Find $\dim R(T)$ and $\dim N(T)$.

My beginning efforts:

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Why don't you write down a typical element of the range space and take a guess at what the range space is. Then figure the dimension. Then use the rank-nullity theorem. – copper.hat Apr 28 '13 at 1:46
Any thoughts about the answers that have been posted? – Gerry Myerson May 6 '13 at 13:51
up vote 1 down vote accepted

$R(T)$ is the column space, now it is $\ {\rm span}\pmatrix{1\\1\\1}$, and has dimension $1$.

Another definition of $R(T)$ is $R(T)=\{Tv\,\mid\,v\in\Bbb R^3\}$, but if you check -in general- the result for $T\pmatrix{1\\0\\0}$, $T\pmatrix{0\\1\\0}$ and $T\pmatrix{0\\0\\1}$ you will see that it's the same.

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Your matrix has three copies of the same vector. The rank is equal to the number of linear independent vectors that appear in your matrix (where it doesn't matter, whether you consider row or column vectors). The nullity is $n$ minus the rank, where your matrix is $n\times m$.

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Yes, I got it. Thank you! ^_^ – Trancot May 7 '13 at 0:40

If $A$ is a matrix representing a linear transformation $T$, then the dimension of the nullspace of $T$ is the dimension of the nullspace of $A$. Do you know how to find the dimension of the nullspace of a matrix, using reduction to row-echelon form?

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No... How do you check? – Trancot Apr 28 '13 at 2:03
Look up Gaussian elimination, e.g. on Wikipedia or YouTube. – Earthliŋ Apr 28 '13 at 10:36
Given any matrix $A$, the solutions of $Ax=0$ form a vector space, called the nullspace of $A$, and you can find a basis for that vector space, and the first step is to use elementary row operations to bring the matrix to reduced row-echelon form. Then, from that for, you can read off some basis vectors for the nullspace of $A$. The exact technique is explained in (just about) every introductory Linear Algebra text ever published. – Gerry Myerson Apr 28 '13 at 12:39

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