# Reason for just a 0 vector in a null space of a L.I matrix

1. Jun 1, 2015

### Akshay Gundeti

Hello Everyone,

Can someone explain why do matrices with linearly independent columns have only 0 vector in their null space?

Thanks

2. Jun 1, 2015

### Svein

Let A be the matrix and $e_{0} = \begin{pmatrix} 1 \\ 0\\ .. \\ 0 \\ \end{pmatrix}$,... $e_{n} = \begin{pmatrix} 0 \\ 0\\ .. \\ 1 \\ \end{pmatrix}$. Then Ae0 is the first column in A, ... and Aen is the last column in A. Therefore, the images of the basis vectors are linearly independent.

3. Jun 1, 2015

### HallsofIvy

Staff Emeritus
This is true IF you are talking about square matrices. The null space of the matrix $$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$$ is a one-dimensional subspace of $R^3$ spanned by $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$.
(It suddenly occurs to me that the columns here are NOT independent so that this is not a good example.)

Last edited: Jun 1, 2015
4. Jun 2, 2015

### Svein

If the number of rows are less than the number of columns, the columns cannot be linearly independent (if there are m rows, with m<n, the maximum number of linearly independent columns are m).

5. Jun 3, 2015

### Hawkeye18

If $\mathbf{a}_1, \mathbf a_2, \ldots \mathbf a_n$ are columns of the matrix $A$, and $\mathbf x=(x_1, x_2, \ldots x_n)^T$, then $$A\mathbf x = \sum_{k=1}^n x_k \mathbf a_k.$$ Then you immediately see from the definition of linear independence that the equation $A\mathbf x = \mathbf 0$ has unique solution if and only if the vectors $\mathbf{a}_1, \mathbf a_2, \ldots \mathbf a_n$ are linearly independent.