# null space of a matrix for a matrix with a 0 column

I'm not a mathematician. Math is a hobby and currently I'm learning bits and pieces of linear algebra when I have time. Confused about a null space for the following matrix

\begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix}

How does one find a null space for this matrix? As far as I understand I need to a such values so that the total would equal to $0$. So what I'm getting is

$$N = \alpha \begin{bmatrix} 0 \\ 0\end{bmatrix} + 0 \begin{bmatrix} 0 \\ 2\end{bmatrix}$$

Wouldn't a coefficient of $0$ make it a trivial solution?

We find the null space by inspection. (Systematic methods are demonstrated in the linked pages below)

Given $$\mathbf{A} = \left( \begin{array}{cc} 0 & 0 \\ 0 & 2 \\ \end{array} \right)$$ we note that $$\mathbf{A} \left( \begin{array}{cc} 1 \\ 0 \\ \end{array} \right) = \left( \begin{array}{cc} 0 \\ 0 \\ \end{array} \right)$$ Both $\color{red}{null}$ spaces are the same for your symmetric matrix: $$\color{red}{\mathcal{N}\left( \mathbf{A} \right)} = \color{red}{\mathcal{N}\left( \mathbf{A}^{*} \right)} =\text{span } % \left\{ \, \color{red}{\left( \begin{array}{cc} 1 \\ 0 \\ \end{array} \right)} \, \right\} %$$

Consider the linear system \begin{align} \mathbf{A} x & = b \\ % \left( \begin{array}{cc} 0 & 0 \\ 0 & 2 \\ \end{array} \right) % \left( \begin{array}{c} x_{1} \\ x_{2} \\ \end{array} \right) % &= \left( \begin{array}{c} b_{1} \\ b_{2} \\ \end{array} \right) \end{align} and specify the data vector to classify the existence and uniqueness of solutions.

Existence and uniqueness

When the data vector has the form $$b = \left( \begin{array}{c} 0 \\ b_{2} \\ \end{array} \right)$$ where $b_{2} \ne 0$, the solution exists and is unique. In fact, the solution is $$x= \color{blue}{\left( \begin{array}{c} 0 \\ \frac{1}{2} b_{2} \\ \end{array} \right)}$$ Existence without uniqueness

When the data vector has the form $$b = \left( \begin{array}{c} b_{1} \\ b_{2} \\ \end{array} \right)$$ where $b_{1} \ne 0$, and $b_{2} \ne 0$, the solution exists and is not unique. The least squares solution is $$x_{LS} = \color{blue}{\left( \begin{array}{c} 0 \\ \frac{1}{2} b_{2} \\ \end{array} \right)} + \alpha \color{red}{\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)}, \qquad \alpha\in \mathbb{C}$$ No existence

When the data vector has the form

$$b = \color{red}{\left( \begin{array}{c} b_{1} \\ 0 \\ \end{array} \right)}$$ where $b_{1} \ne 0$, the solution does not exist.