# Finding the basis of the null space and the image of a matrix

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?

• 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. Feb 25, 2020 at 19:02
• So my matrix is of rank 3 and the first 3 columns of my original matrix are the basis of its image? Feb 25, 2020 at 19:11

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