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

• Akshay Gundeti
In summary, matrices with linearly independent columns will have only the zero vector in their null space. This is because the images of the basis vectors are linearly independent, and this holds true for square matrices. If the number of rows is less than the number of columns, the columns cannot be linearly independent. This can be seen through the equation A*x = 0, where A is the matrix and x is a vector composed of coefficients.
Akshay Gundeti
Hello Everyone,

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

Thanks

Akshay Gundeti said:
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 Aen is the last column in A. Therefore, the images of the basis vectors are linearly independent.

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 by a moderator:
HallsofIvy said:
(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).

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.

What is a null space of a L.I matrix?

A null space, also known as a kernel, is the set of all vectors that when multiplied by a matrix result in a zero vector. A L.I matrix, or a linearly independent matrix, is a matrix where the columns are not linear combinations of each other.

Why does a L.I matrix have a null space?

A L.I matrix has a null space because it contains at least one linearly independent column that cannot be represented as a linear combination of the other columns. This results in the existence of a vector that when multiplied by the matrix, produces a zero vector.

What does a zero vector in the null space of a L.I matrix represent?

The zero vector in the null space of a L.I matrix represents a vector that is mapped to the origin when multiplied by the matrix. In other words, it is a vector that does not contribute to the output of the matrix multiplication.

Can a L.I matrix have more than one zero vector in its null space?

No, a L.I matrix can only have one zero vector in its null space. This is because if there were more than one zero vector, it would mean that there are multiple vectors that are mapped to the origin when multiplied by the matrix, which would contradict the definition of a null space.

How does the null space of a L.I matrix relate to its invertibility?

The null space of a L.I matrix is directly related to its invertibility. A L.I matrix is invertible if and only if its null space only contains the zero vector. This means that the matrix does not map any non-zero vector to the origin, making it possible to obtain a unique inverse for the matrix.

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