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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?

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  • $\begingroup$ if ${\rm rank}A=3$ it means that you have an injective linear map $\Bbb{R}^3\to\Bbb{R}^4$ $\endgroup$
    – janmarqz
    Commented Jan 9, 2014 at 0:43
  • $\begingroup$ The vector space $\{0\}$ is a perfectly valid kernel. $\endgroup$ Commented Feb 22, 2023 at 17:11

1 Answer 1

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The rank theorem is optimal.

Otherwise just solve it like you would with any other numbers:

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.

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