# kernel matrix with trivial solution only

When a matrix have only the trivial solution (zero vertor). How i must represent the solve?

For a particular case of matix 4x3 $$A = \begin{bmatrix} 1 & 1 & -3 \\ 0 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & -1 & -4 \end{bmatrix}$$

After do some operations get its reduced form. (R3-R1, R4-R1), (R4+R2, R2-2R3), (-R2/7), (R1+3R2, R3-4R2), (R1-R3), (R3<->R2) $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} With\ Rank(A) = 3$$

I found diferent cases for the solve

1. First (I think is wrong)

$$ker(A) = A\vec{x} = \nexists$$

1. Second

$$ker(A) = A\vec{x} = \begin{Bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} \end{Bmatrix}$$

For the rank theorem $$nul(A) = dim(ker(A)) = n - rank(A)= 3 - 3 = 0$$

In adition, this solution satisfy the properies of a subspace:

• Contains the null element of the space
• Closed under the sum of element and scalar product

Any idea what is the right solve?

• if ${\rm rank}A=3$ it means that you have an injective linear map $\Bbb{R}^3\to\Bbb{R}^4$ Jan 9, 2014 at 0:43
• The vector space $\{0\}$ is a perfectly valid kernel. Feb 22 at 17:11

if the vectors are represented by $x, y, z$, then your system becomes
$$x = 0; y = 0; z = 0; 0=0$$ which has as unique solution the null vector.