Let $A=\begin{bmatrix} a & b\\a& b\end{bmatrix}$ where $a>0$. Find rank $A$. Find a basis of the null space and a basis of the row space of $A$; and show that they are perpendicular.
My attempt at a solution:
I'm assuming that any matrix in $A$ is of the form $x=\alpha\begin{bmatrix}1 &0\\
1&0\end{bmatrix}$. From this, rank $A=1$, and a basis of the row space is $([1,0])$. However, I don't know how to find a basis of the null space in this particular problem.
Is the assumption that any matrix in $A$ is of the form $x=\alpha\begin{bmatrix}1 &0\\
1&0\end{bmatrix}$ right? Since I can't find a basis for the null space, I don't know if my assumption is right. In this case, can the null space be the zero vector?
Best Answer
HINT
To start with, notice the columns are proportional. So that $\text{rank}(A) = 1$.
In order to find out a basis for the kernel, you have to solve the system of linear equations: \begin{align*} \begin{bmatrix} a & b\\ a & b \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} & \Longleftrightarrow ax + by = 0 \Longleftrightarrow x = -\frac{by}{a} \end{align*}
Hence we conclude that $\ker(A) = \text{span}\left\{\left(-\dfrac{b}{a},1\right)\right\}$, whose dimension equals one.
Similarly, in order to find a basis for the row space, notice the rows are LD.
Consequently, we conclude the row space equals $\text{span}\left\{\left(1,\dfrac{b}{a}\right)\right\}$.
Based on such results, can you take it from here?