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I need to know if my work is right.

T$\begin{bmatrix}a11 & a12 & a13\\a21 & a22 & a23\\a31 & a32 & a33\end{bmatrix}$=$\begin{bmatrix}a11+a12+a13 & a21+a22+a23\\ a31+a32+a33 & 0\end{bmatrix}$

Question: Find bases for the nullity and rank

Attempt:

Basis for N(T)={(-1,0,1), (-1,1,0), (-1,0,1),(-1,1,0), (-1,0,1), (-1,1,0) } Dim(N(T))=6

To find the rank, I used the dimension theorem. Dim V=nullity T + Rank T.

Dim V=9 and Nullity T= 6 so 9-6= Rank T = 3

Is the basis for the range just the standard basis for M2x2?

Essie
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  • $\begin{bmatrix} 0&0\0&1\end{bmatrix}$ is not in the image of $T.$ And, you know that the image of $T$ is a 3 dimensional space. – Doug M Apr 13 '18 at 22:45
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    Terminology nitpicking: nullity and rank are numbers. The kernel and image of the transformation are vector spaces and have bases. The dimensions of these spaces are the nullity and rank, respectively. – amd Apr 13 '18 at 22:52
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    How is a set of 3-element vectors a basis for the null space of $T$ when its domain is $3\times3$ matrices? – amd Apr 13 '18 at 22:54
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    A suggestion: in case if the fact that the transformation is between two matrix spaces that is confusing you, consider the essentially identical question of the transformation from $\Bbb R^9$ to $\Bbb R^4$ such that $T(a,b,c,d,e,f,g,h,i)=(a+b+c,d+e+f,g+h+i,~~0)$ – JMoravitz Apr 13 '18 at 23:12
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    Given your attempt, it appears that perhaps you had the right vague idea of what to do, but was unable to properly express it. Using the reformulation of the question from my earlier comment, your basis could be $(-1,0,1,0,0,0,0,0,0), (-1,1,0,0,0,0,0,0,0),(0,0,0,-1,0,1,0,0,0),\dots$. Your error, as pointed out, is that you seemingly wrote only the nonzero row of the basis matrices, which makes your basis not only seem redundant (you wrote (-1,0,1) three times for example), but makes them appear to not even be elements of the domain. As for the range, note the fourth entry is always zero. – JMoravitz Apr 13 '18 at 23:19
  • So dim N(T)=2 ? and a basis for the range is ( 1,0,0,0) , (0,1,0,0) (0,0,1,0) ? – Essie Apr 13 '18 at 23:44
  • That basis is fine for the range of the transformation from $\Bbb R^9$ to $\Bbb R^4$., but again it is not fine for the transformation from $M_{3\times 3}$ to $M_{2\times 2}$. The dimension is not $2$ however... remember that the dimension is the number of vectors in the respective basis. – JMoravitz Apr 14 '18 at 04:56

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