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
Best Answer
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):
$$\begin{bmatrix} 1\\3\\-1 \end{bmatrix}, \begin{bmatrix} 1\\-1\\2 \end{bmatrix}, \begin{bmatrix} -1\\2\\-4\\ \end{bmatrix}. $$