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Given $A = \begin{bmatrix} 1&1&-1&0 \\3&-1&2&-4 \\-1&2&-4&3 \end{bmatrix}$ I need to calate the following:

1) Find a basis for the null space and the nullity of A.
2) Find a basis for the row space and the row rank of A.
3) Find a basis for the column space and the column rank of A.

So I brought the matrix $A$ in row echelon form:

$\begin{bmatrix} 1&1&-1&0 \\0&1&0&0 \\0&0&1&0 \end{bmatrix}$ I believe I did this part correctly.

So I know the rank of this matrix $A$ is 3 and it has 3 pivot points. Looking for some help with these questions

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  • $\begingroup$ You are correct, this matrix is of rank $3$. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Oct 7, 2016 at 12:58
  • $\begingroup$ Do you already know the rank-nullity theorem, also known (perhaps in a little more advanced degree) as the dimensions theorem? $\endgroup$
    – DonAntonio
    Commented Oct 7, 2016 at 12:59
  • $\begingroup$ rank + nullity = n is a formula i know $\endgroup$
    – user376097
    Commented Oct 7, 2016 at 13:00
  • $\begingroup$ How do i go about solving these problems? $\endgroup$
    – user376097
    Commented Oct 7, 2016 at 13:00

1 Answer 1

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Why not just reduce the matrix ? its not more work and often makes things easier.

\begin{bmatrix} 1&0&0&0 \\0&1&0&0 \\0&0&1&0 \end{bmatrix}

1) We see that the row and column rank (they are always equal) are $3$. The nullity $=n-$rank$=4-3=1$. To find the null space we have to solve $A\mathbf{x}=0$, and this is easy now in row reduced form $\mathbf{x}=\begin{bmatrix} 0\\0\\0\\a \end{bmatrix}$.

2) The row space has dimension $3$ as mentioned, for the basis one can take: $$(1,0,0,0)$$ $$(0,1,0,0)$$ $$(0,0,1,0)$$

Or one could take the rows of the original matrix, since the rank is $3$.

3) The column rank is also $3$ row reduction has not changed the column vectors, just expressed them in a different basis so a basis for the column space will be the first $3$ vectors of the original matrix (corresponding to the pivot position of the reduced matrix):

$$\begin{bmatrix} 1\\3\\-1 \end{bmatrix}, \begin{bmatrix} 1\\-1\\2 \end{bmatrix}, \begin{bmatrix} -1\\2\\-4\\ \end{bmatrix}. $$

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  • $\begingroup$ did you just take my reduced matrix and reduce the first row even farther? $\endgroup$
    – user376097
    Commented Oct 7, 2016 at 13:09
  • $\begingroup$ Yes, and it wasnt any work at all ! $\endgroup$
    – Rene Schipperus
    Commented Oct 7, 2016 at 13:12
  • $\begingroup$ awesome thanks for your time, that makes things easier when reduced like that $\endgroup$
    – user376097
    Commented Oct 7, 2016 at 13:13
  • $\begingroup$ @ Rene Schipperus for b) is that supposed to be a matrix or is it supposed to be in the form that you wrote it? $\endgroup$
    – user376097
    Commented Oct 7, 2016 at 13:17
  • $\begingroup$ Do you mean for 2) ? That is three vectors. Not sure I understand what your asking. $\endgroup$
    – Rene Schipperus
    Commented Oct 7, 2016 at 13:18

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