Significance of adding zero rows and columns to a matrix. Suppose I have a $n\times n$ matrix to which I add a row of zeros and a column of zeros, somewhere in the matrix, to make it $(n+1)\times (n+1)$. When I multiply the matrix by itself, or multiply it with other matrices where I have inserted a row and column of zeros in the same way, it seems to behave as though the extra row/column were not there. Is there a way to think about the added row/column, and why it does not affect the product? What have I done to this matrix? I think (but not sure at all) that it is like I have put my matrix in a higher matrix space but it only spans the subset where it previously existed?
 A: It's like being confined to a plane. The vectors all have length $2$. When you add a row of $0$s, you add a $\color{red}{null}$ space. The inhabitants of the plane can't see any change because the addition is a $\color{red}{null}$ space.

No null spaces
$$
\mathbf{A} =
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right)
\in\mathbb{C}^{2\times2}_{2}
$$
Life is good. The linear system 
$$
 \mathbf{A} x = b
$$
always has a unique solution
$$
 x = b
$$
Add a null space
Add a row of $0$s.
$$
\mathbf{A} =
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
 0 & 0 \\
\end{array}
\right)
\in\mathbb{C}^{3\times2}_{2}
$$
Life isn't so good. We have solvability conditions based on the subspace decomposition of the data
$$
b =
\left(
\begin{array}{c}
 b_{1} \\
 b_{2} \\
 b_{3} \\
\end{array}
\right)
=
\color{blue}{b_{\mathcal{R}}} + 
\color{red}{b_{\mathcal{N}}}
=
\color{blue}{\left(
\begin{array}{c}
 b_{1} \\
 b_{2} \\
 0 \\
\end{array}
\right)}
+
\color{red}{\left(
\begin{array}{c}
 0 \\
 0 \\
 b_{3} \\
\end{array}
\right)}
$$


*

*If $b_{1} = b_{2} = 0$, there is no solution.

*If $b_{1} \ne 0$, or $b_{2} \ne 0$, and $b_{3} = 0$ a unique solution exists.

*If $b_{1} \ne 0$, or $b_{2} \ne 0$, and $b_{3} = 0$ there are an infinite number of solutions.

Fundamental Theorem of Linear Algebra
The four fundamental subspaces for an arbitrary matrix $\mathbf{A}\in\mathbb{C}^{m\times n}_{\rho}$ can be expressed as
$$
\begin{align}
%
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
    \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
  \mathbf{C}^{m} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
    \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
\end{align}
$$
The game you are playing involves one of these two themes:
$$
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} 
    \quad \Rightarrow \quad 
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
    \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
$$
or
$$
  \mathbf{C}^{m} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} 
    \quad \Rightarrow \quad 
    \mathbf{C}^{m} =
      \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
      \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} \\
$$

Explore MSE

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