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I have the following matrix:

\begin{bmatrix}1&0&0&3\\0&2&1&2\\-1&-1&0&-2\\-1&0&1&1\end{bmatrix}

I want to find the basis of its null space and of its image. As far as I understand, the null space is the set of all non-zero vectors that produce a $0$ when multiplying this matrix ($Ax = 0$). An image of this matrix would be the set of all vectors I can get by multiplying this matrix by a vector.

I've reduced the matrix to row echelon form: \begin{bmatrix}1&0&0&3\\0&1&0&-1\\0&0&1&4\\0&0&0&0\end{bmatrix} From this I get the equations: $x_1 = -3x_4, x_2 = x_4, x_3 = -4x_4$. This is where I got stuck. Does the basis of the null space only contain the vector $[-3, 1, -4, 1]$? How do I go about finding the basis for the image?

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  • $\begingroup$ Yes. As for your second question, can you figure out what the rank of the matrix is? The image of a matrix is spanned by any $n$ of its linearly independent columns, where $n$ is the rank of the matrix. $\endgroup$ Commented Feb 25, 2020 at 19:02
  • $\begingroup$ So my matrix is of rank 3 and the first 3 columns of my original matrix are the basis of its image? $\endgroup$ Commented Feb 25, 2020 at 19:11

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Your calculation for the basis of the kernel is correct.

Concerning your second question: Since the matrix has rank 3, you need three linearly independent column vectors of the matrix as a basis for the image. You can take the first three vectors, since they are linearly independent. Therefore, a basis of the image is

$$(1, 0, -1, -1), \qquad (0, 2, -1, 0), \qquad (0, 1, 0, 1).$$

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  • $\begingroup$ Should I not take the first three columns of the original matrix, rather the columns of the reduced form? $\endgroup$ Commented Feb 25, 2020 at 19:53
  • $\begingroup$ Ah, I looked at the wrong matrix, of course. Edited, thanks. $\endgroup$ Commented Feb 25, 2020 at 19:57

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