My Approach:
For convenience I am taking an example to prove this statement,
Let $$A=\begin{bmatrix} 1&0&1&2 \\2&3&1&1 \\1&2&3&1 \end{bmatrix} $$ $\implies A:F^4\to F^3$
Null space of $A$ is defined as:
$$N_A=\{ \vec{x} \in F^4 : A \vec{x}=0 \}$$
Let $\vec{x}= \begin{bmatrix} x_{1} \\ x_{2} \\ x_3 \\ x_4 \end{bmatrix}$
$\implies$ $$ x_1 + x_3+2x_4=0$$ $$ 2x_1 +3x_2+ x_3+x_4=0$$ $$ x_1 +2x_2+ 3x_3+x_4=0$$ Which on solving gives $$\vec{x}= x_4\begin{bmatrix} -13\\ 7 \\ -3\\ 8 \end{bmatrix} = k\begin{bmatrix} -13\\ 7 \\ -3\\ 8 \end{bmatrix} ,\forall k\in R$$ Now,Row space of $A$ is defined as: $$\text{Row} (A) = \{ \vec{r} \in F^4 : \vec{r}=b_1 \vec{r_1} + +b_2 \vec{r_2}+b_3 \vec{r_3}\}$$ Where, $\vec{r_1}= \begin{bmatrix} 1\\ 0 \\ 1\\ 2 \end{bmatrix}$ , $\vec{r_2}= \begin{bmatrix} 2\\ 3 \\ 1\\ 1 \end{bmatrix}$ and $\vec{r_3}= \begin{bmatrix} 1\\ 2 \\ 3\\ 1 \end{bmatrix}$
Now, if we can show that $\vec{x} $ can be represented as a linear combinations of $\vec{r_1},\vec{r_2 }\& \vec{r_3}$ , then we have proved that $\vec{x} \in $ Row( $A$)
i.e, R.T.P $\to$ $$ \vec{x}= d_1 \vec{r_1} + +d_2 \vec{r_2}+d_3 \vec{r_3}$$
$\implies$ $$-13k= d_1 + 2d_2+d_3$$ $$7k= 3d_2+2d_3$$ $$-3k= d_1 + d_2+3d_3$$ $$8k= 2d_1 + d_2+d_3$$ On solving, we find there doesn't exist any $d_1,d_2 \& d_3$ which will satisfy these four equations simultaneously

So, where I have done the mistake? Any help please...

  • $\begingroup$ Are you sure that you were meant to prove that the two spaces are the same? They are always, as the answer below points out, orthogonal complements, so they will never be identical for any matrix. $\endgroup$ – amd Dec 8 '19 at 21:57
  • $\begingroup$ @amd Actually I was taught in my academics that Range space and column space are equal subspaces of a matrix ; similarly Null space and Row space are equal subspaces; 1st statement i have proved but i was facing difficulty in proving the next statement; that's why asked here... $\endgroup$ – Suresh Dec 9 '19 at 5:03
  • $\begingroup$ Nope. The null space and row space are never equal. A simple counterexample is any full-rank square matrix, say, the identity matrix: its null space contains only the zero vector. That’s why you haven’t been able to prove that they are. One of the key relationships among the four fundamental spaces of a matrix is that they are orthogonal complements. It appears that you were either taught incorrectly or have gotten something confused. $\endgroup$ – amd Dec 9 '19 at 5:34

The row space is obviously not equal to the null space: any row of $A$ is in the row space, but a non-zero row (transposed) is not in the null space, as a non-zero vector cannot have an inner product with itself that is $0$.

$x$ is in the null space of $A$ iff it is orthogonal to the row space of $A$ iff it is in the column space of $A^T$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.