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

[Akshay Gundeti]

Hello Everyone,

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

Thanks

 

[Svein]

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

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 Ae_n is the last column in A. Therefore, the images of the basis vectors are linearly independent.

 

[HallsofIvy]

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.)

 

[Svein]

(It suddenly occurs to me that the columns here are NOT independent so that this is not a good example.)

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).

 

[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.