Four Fundamental Subspaces

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In this article we study the four fundamental subspaces

  • Column Space $C(\mathbf{A})$ in $\mathbb{R}^m$
  • Nullspace $N(\mathbf{A})$ in $\mathbb{R}^{n-r}$
  • Row space $C(\mathbf{A}^T)$ in $\mathbb{R}^{n}$
  • Left Nullspace $N(\mathbf{A}^T)$ in $\mathbb{R}^{m-r}$
The dimensions of The Nullspace and the row space add to $n$. The dimension of the Column Space and the Left Nullspace add to $m$.

We have already learned about the Column Space and the Nullspace in other articles. Also it is immediately clear that the row space can be written as the column space of $\mathbf{A}^T$. But we should have a closer look on the Left Nullspace.

\begin{align} \mathbf{A}^T \mathbf{y} = \mathbf{0} \end{align}

If we transpose this equation $\mathbf{y}^T$ multiplies the matrix $\mathbf{A}$ from the left \begin{align} \mathbf{y}^T \mathbf{A} = \mathbf{0}^T \end{align} we thus call the associated Nullspace the left Nullspace.


Example


To elaborate on the subspaces we study the following matrix $m \times n = 3 \times 4$ \begin{align} \begin{bmatrix} {1} & 2 & 3 & 1\\ 1 & 1 & 2 & 1\\ {1} & 2 & 3 & 1 \end{bmatrix} \end{align} Doing Matrix Elimination we have \begin{align} \begin{array}{rrrr} {1} & 2 & 3 & 1\\ 1 & 1 & 2 & 1\\ {1} & 2 & 3 & 1 \end{array} \rightarrow \begin{array}{rrrr} {1} & 2 & 3 & 1\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 \end{array} \end{align} where the matrix is upper triangular $\mathbf{U}$ \begin{align} \mathbf{U}=\begin{bmatrix} {1} & 2 & 3 & 1\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{align} cleaning up above the pivots we get \begin{align} \mathbf{R}=\begin{bmatrix} {1} & 0 & 1 & 1\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}=\begin{bmatrix} \mathbf{I} & \mathbf{F}\\ \mathbf{0} & \mathbf{0} \end{bmatrix} \end{align} Notice, we did row operations on the matrix, so the row space stays conserved, but no the column space $C(\mathbf{A}) \not= C(\mathbf{R})$. The third row of $\mathbf{R}$ is all zeros, so the rank $r=2$ and the first two rows of $\mathbf{R}$ form a basis for the row space, so $dim [C(\mathbf{A}^T)]=r$. If we write the steps of Matrix Elimination in matrix form we can write elimination as \begin{align} \mathbf{E}\mathbf{A}&=\mathbf{R}\\ \begin{bmatrix} {-1} & 2 & 0 \\ 1 & -1 & 0 \\ {-1} & 0 & 1 \end{bmatrix} \begin{bmatrix} {1} & 2 & 3 & 1\\ 1 & 1 & 2 & 1\\ {1} & 2 & 3 & 1 \end{bmatrix} &= \begin{bmatrix} {1} & 0 & 1 & 1\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{align} as we discussed in the article on LU Decomposition. The last row of $\mathbf{E}$ produces a row of zeros, so it is a vector in the left Nullspace. In fact, $\begin{bmatrix} {-1} & 0 & 1 \end{bmatrix}$ is the only vector in the left Nullspace and it forms a basis for this left Nullspace ($dim[N(\mathbf{A}^T)]=m-r=3-2=1$).




Video Lectures:

  • Gilbert Strang - Introduction to Linear Algebra Lec. 10