Question

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

Answer 1

Step 1
Given:
\(A=\begin{bmatrix}3 & 4 \\6 & 8 \end{bmatrix}\)
Step 2
\(A^T=\begin{bmatrix}3 & 6\\4 & 8 \end{bmatrix}\)
Apply \(R_2 \rightarrow R_2-\frac{4}{3}R_1\)
\(=\begin{bmatrix}3 & 6 \\0 & 0 \end{bmatrix}\)
whose rank is 1
\(R(A^T)=\text{Span}\left\{\begin{pmatrix}3 \\4 \end{pmatrix}\right\}\)
Now, \(N(A^T)\)
\(3x+6y=0\)
\(\Rightarrow x=-2y\)
\(\begin{pmatrix}x \\y \end{pmatrix}=\begin{pmatrix}-2y \\y \end{pmatrix}\)
\(=\begin{pmatrix}-2 \\1 \end{pmatrix} \ \ [\therefore y=1]\)
\(N(A^T)=\text{Span}\left\{\begin{pmatrix}-2 \\1 \end{pmatrix}\right\}\)
\(A=\begin{bmatrix}3 & 4\\6 & 8 \end{bmatrix}\)
Apply \(R_2 \rightarrow R_2-2R_1\)
\(\begin{bmatrix}3 & 4\\0 & 0 \end{bmatrix}\)
Whose rank is 1
\(R(A)=\text{Span}\left\{\begin{pmatrix}3 \\6 \end{pmatrix}\right\}\)
For N(A)
\(3x+4y=0\)
\(\Rightarrow x=-\frac{4}{3}y\)
Step 3
\(\begin{pmatrix}x \\y \end{pmatrix}=\begin{pmatrix}-\frac{4}{3}y \\y \end{pmatrix}\)
\(=\begin{pmatrix}-\frac{4}{3} \\ 1 \end{pmatrix} \ \ [\therefore y=1]\)
\(N(A)=\text{Span}\left\{\begin{pmatrix}-\frac{4}{3} \\ 1 \end{pmatrix}\right\}\)