Linear Algebra – Matrices – Subspaces

Definition: A subset \(H\) of \(\mathbb{R}^n\) is called a subspace of \(\mathbb{R}^n\) if:

1. \(\mathbf{0}\in H\);


2. \(\mathbf{u}+\mathbf{v}\in H\) for all \(\mathbf{u},\mathbf{v}\in H\);


3. \(c\mathbf{u}\in H\) for all \(\mathbf{u}\in H\) and all \(c\in\mathbb{R}\).

The first condition prevents the set \(H\) from being empty. If the set \(H\) is not empty, then there exists at least one vector in \(H\). Then, by the third condition we can multiply this vector by the number \(0\) which implies that \(\mathbf{0}\in H\). We als have that the opposite of the vector is in \(H\) (take \(c=-1\) in the third condition) and then we have by the second condition that the sum of these two vectors, and that is the zero vector, is in \(H\) too.

The first condition comes in handy, because this immediately implies that every subset not containing the zero vector cannot be a subspace.

Example: If \(\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_p\) are vectors in \(\mathbb{R}^n\), then the set of all linear combinations of \(\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_p\) is a subspace of \(\mathbb{R}^n\).

Notation: \(\text{Span}\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_p\}\) is the subspace spanned (or generated) by \(\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_p\).

Note that \(\mathbb{R}^n\) itself is a subspace of \(\mathbb{R}^n\). Another special case is the set \(\{\mathbf{0}\}\); this is also a subspce of \(\mathbb{R}^n\). These subspaces are sometimes called trivial subspaces.

Definition: The column space of a matrix \(A\) is the set \(\text{Col}(A)\) of all linear combinations of the columns of \(A\).

Definition: The null space of a matrix \(A\) is the set \(\text{Nul}(A)\) of all solutions of the homogeneous matrix equation \(A\mathbf{x}=\mathbf{0}\).

Theorem: If \(A\) is an \(m\times n\) matrix, then \(\text{Col}(A)\) is a subspace of \(\mathbb{R}^m\) and \(\text{Nul}(A)\) is a subspace of \(\mathbb{R}^n\).

Definition: A basis of a subspace \(H\) of \(\mathbb{R}^n\) is a linear independent set that spans \(H\).

Theorem: The pivot columns of a matrix \(A\) form a basis of the column space \(\text{Col}(A)\).

Example: Consider the matrix \(A=\begin{pmatrix}1&0&-1\\-1&1&2\\2&-1&-3\end{pmatrix}\), then we have:

\[A=\begin{pmatrix}1&0&-1\\-1&1&2\\2&-1&-3\end{pmatrix}\sim\begin{pmatrix}1&0&-1\\0&1&1\\0&-1&-1\end{pmatrix} \sim\begin{pmatrix}1&0&-1\\0&1&1\\0&0&0\end{pmatrix}.\]

This implies that for instance \(\left\{\begin{pmatrix}1\\-1\\2\end{pmatrix},\begin{pmatrix}0\\1\\-1\end{pmatrix}\right\}\) is a basis of the column space \(\text{Col}(A)\). Further we have for \(A\mathbf{x}=\mathbf{0}\) that \(\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\) with \(\left\{\begin{array}{l}x_1=x_3\\x_2=-x_3\\x_3\;\text{is free.}\end{array}\right.\quad\) Hence: \(\mathbf{x}=\begin{pmatrix}x_3\\-x_3\\x_3\end{pmatrix} =x_3\begin{pmatrix}1\\-1\\1\end{pmatrix}\) and this implies that \(\left\{\begin{pmatrix}1\\-1\\1\end{pmatrix}\right\}\) is a basis of the null space \(\text{Nul}(A)\).

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Last modified on March 29, 2021