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 <br /> e_{0} = \begin{pmatrix}<br /> 1 \\<br /> 0\\<br /> .. \\<br /> 0 \\<br /> \end{pmatrix}<br />,... <br /> e_{n} = \begin{pmatrix}<br /> 0 \\<br /> 0\\<br /> .. \\<br /> 1 \\<br /> \end{pmatrix}<br />. 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 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 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.