For each of the following matrices, determine a basis for each of the subspaces R(AT), N(A), R(A), and N(AT): A=begin{bmatrix}3 & 4 6 & 8 end{bmatrix}

Question

For each of the following matrices, determine a basis for each of the subspaces R(AT), N(A), R(A), and N(AT):

A = [\begin{matrix} 3 & 4 \\ 6 & 8 \end{matrix}]

Answer 1

Step 1
Given:
$A = [\begin{matrix} 3 & 4 \\ 6 & 8 \end{matrix}]$
Step 2
$A^{T} = [\begin{matrix} 3 & 6 \\ 4 & 8 \end{matrix}]$
Apply $R_{2} \to R_{2} - \frac{4}{3} R_{1}$
$= [\begin{matrix} 3 & 6 \\ 0 & 0 \end{matrix}]$
whose rank is 1
$R (A^{T}) = (\begin{matrix} 3 \\ 4 \end{matrix})$
Now, $N (A^{T})$
$3 x + 6 y = 0$
$\Rightarrow x = - 2 y$
$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - 2 y \\ y \end{matrix})$
$= (\begin{matrix} - 2 \\ 1 \end{matrix}) [∴ y = 1]$
$N (A^{T}) = (\begin{matrix} - 2 \\ 1 \end{matrix})$
$A = [\begin{matrix} 3 & 4 \\ 6 & 8 \end{matrix}]$
Apply $R_{2} \to R_{2} - 2 R_{1}$
$[\begin{matrix} 3 & 4 \\ 0 & 0 \end{matrix}]$
Whose rank is 1
$R (A) = (\begin{matrix} 3 \\ 6 \end{matrix})$
For N(A)
$3 x + 4 y = 0$
$\Rightarrow x = - \frac{4}{3} y$
Step 3
$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} - \frac{4}{3} y \\ y \end{matrix})$
$= (\begin{matrix} - \frac{4}{3} \\ 1 \end{matrix}) [∴ y = 1]$
$N (A) = (\begin{matrix} - \frac{4}{3} \\ 1 \end{matrix})$