# Math Help - Determine a basis for a subspace

1. ## Determine a basis for a subspace

For the following matrix, determine a basis for each of the subspace $R(A^{\top})$, $N(A)$, $R(A)$, and $N(A^{\top})$.

A=\left[\begin {array}{cc}3&4\\\noalign{\medskip}6&8\end{array}\r ight]

So I found the RREF of A to be \left[\begin {array}{cc}1&\dfrac{4}{3}\\\noalign{\medskip}0&0\e nd{array}\right]
$(1,\dfrac{4}{3})$ forms a basis for $R(A)$ and \left[\begin {array}{c}1\\\noalign{\medskip}\dfrac{4}{3}\end{ar ray}\right] forms a basis for $R(A^{\top})$.
$x_{1}+\dfrac{4}{3}x_{2}=0$
Setting $x_{1}=\alpha$ and $x_{2}=\beta$ I got
\textbf{x}=\alpha\left[\begin {array}{c}0\\\noalign{\medskip}\dfrac{-3}{4}\end{array}\right]+\beta\left[\begin {array}{c}\dfrac{-4}{3}\\\noalign{\medskip}0\end{array}\right]
So x is a equal to the N(A)? This is where I am stuck, I don't know where to go from here or even if I begun my process correctly. Help would be greatly appreciated, thank you!

2. What is $N(A)$? I know $R(A)$ is the row space of $A$.

3. Originally Posted by KGene
What is $N(A)$? I know $R(A)$ is the row space of $A$.
$N(A)$ is the nullspace of A.

4. Originally Posted by pakman
$N(A)$ is the nullspace of A.
Suppose $x=(x_1,x_2)$ is in the nullspace of $A$. By looking at the row space, it must satisfies the equation $x_1+4/3x_2=0$ as you mentioned earlier. So $x_1,x_2$ are related as given in the equation. Can you write $x$ in term of $x_1$ only?
If you could do so, factor out $x_1$, then you can easily see a basis element of $N(A)$.

Hope this helps.

5. Note also that $\dim N(A)=2-\dim R(A)=2-1=1$.

6. Hi !

Originally Posted by KGene
Suppose $x=(x_1,x_2)$ is in the nullspace of $A$. By looking at the row space, it must satisfies the equation $x_1+4/3x_2=0$ as you mentioned earlier. So $x_1,x_2$ are related as given in the equation. Can you write $x$ in term of $x_1$ only?
If you could do so, factor out $x_1$, then you can easily see a basis element of $N(A)$.

Hope this helps.
Or just find a vector $x=\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ that satisfies this equality (apart from the null vector )

7. Originally Posted by KGene
Suppose $x=(x_1,x_2)$ is in the nullspace of $A$. By looking at the row space, it must satisfies the equation $x_1+4/3x_2=0$ as you mentioned earlier. So $x_1,x_2$ are related as given in the equation. Can you write $x$ in term of $x_1$ only?
If you could do so, factor out $x_1$, then you can easily see a basis element of $N(A)$.

Hope this helps.
So x in terms of only $x_{1}$ would be
\textbf{x}=x1\left[\begin {array}{c}-1\\\noalign{\medskip}\dfrac{-3}{4}\end{array}\right]

Did you mean it like that?

8. Originally Posted by pakman
So x in terms of only $x_{1}$ would be
\textbf{x}=x1\left[\begin {array}{c}-1\\\noalign{\medskip}\dfrac{-3}{4}\end{array}\right]

Did you mean it like that?
There is a mistake to be fixed, $x_2=-3/4x_1$. So it should be $x_1\left [\begin{array}{c}1\\ -3/4\end{array}\right ]$ instead.