TeachingLinear Algebra – Matrices – Dimension and rank
Definition: Suppose that \(\mathcal{B}=\{\mathbf{b}_1,\ldots,\mathbf{b}_p\}\) is a basis of a subspace \(H\). For each vector \(\mathbf{x}\) in \(H\) the coordinates of \(\mathbf{x}\) with respect to the basis \(\mathcal{B}\) are the weights \(c_1,\ldots,c_p\) such that \(\mathbf{x}=c_1\mathbf{b}_1+\cdots+c_p\mathbf{b}_p\) and the vector
\[[\mathbf{x}]_{\mathcal{B}}=\begin{pmatrix}c_1\\\vdots\\c_p\end{pmatrix}\]in \(\mathbb{R}^p\) is called the coordinate vector of \(\mathbf{x}\) with respect to the basis \(\mathcal{B}\).
Example: Consider the subspace \(H\) of \(\mathbb{R}^3\) spanned by \(\mathbf{b}_1=\begin{pmatrix}1\\-1\\2\end{pmatrix}\) and \(\mathbf{b}_2=\begin{pmatrix}2\\-1\\3\end{pmatrix}\). Then \(\{\mathbf{b}_1,\mathbf{b}_2\}\) is a basis of \(H\). For the vector \(\mathbf{x}=\begin{pmatrix}1\\1\\0\end{pmatrix}\) we have that \(\mathbf{x}\in H\), since:
\[c_1\mathbf{b}_1+c_2\mathbf{b}_2=\mathbf{x}:\quad\left(\left.\begin{matrix}1&2\\-1&-1\\2&3\end{matrix}\,\right|\,\begin{matrix}1\\1\\0\end{matrix}\right) \sim\left(\left.\begin{matrix}1&2\\0&1\\0&-1\end{matrix}\,\right|\,\begin{matrix}1\\2\\-2\end{matrix}\right) \sim\left(\left.\begin{matrix}1&0\\0&1\\0&0\end{matrix}\,\right|\,\begin{matrix}-3\\2\\0\end{matrix}\right) \quad\Longrightarrow\quad\left\{\begin{array}{l}c_1=-3\\[2.5mm]c_2=2\end{array}\right.\quad\Longrightarrow\quad[\mathbf{x}]_{\mathcal{B}}=\begin{pmatrix}-3\\2\end{pmatrix}.\]Further we have (for instance) that: \([\mathbf{y}]_{\mathcal{B}}=\begin{pmatrix}-1\\1\end{pmatrix}\quad\Longrightarrow\quad\mathbf{y} =-\mathbf{b}_1+\mathbf{b}_2=-\begin{pmatrix}1\\-1\\2\end{pmatrix}+\begin{pmatrix}2\\-1\\3\end{pmatrix}=\begin{pmatrix}1\\0\\1\end{pmatrix}\).
Theorem: If a basis of a subspace \(H\) consists of \(p\) vectors, then every basis of \(H\) consists of \(p\) vectors.
Definition: The dimension of a subspace \(H\neq\{\mathbf{0}\}\) is the number of vectors in any basis of \(H\). The dimension of the (trivial) subspace \(\{\mathbf{0}\}\) is \(0\) since it has no basis.
Definition: The rank of a matrix \(A\) is the dimension of the column space of \(A\).
Theorem: For a matrix \(A\) with \(n\) colums we have: \(\dim\left(\text{Col}(A)\right)+\dim\left(\text{Nul}(A)\right)=n\).
Proof: We have: \(\dim\left(\text{Col}(A)\right)=\text{rank}(A)\) is equal to the number of pivot columns of \(A\). Further we have: \(\dim\left(\text{Nul}(A)\right)\) is equal to the number of free variables in \(A\mathbf{x}=\mathbf{0}\) and that is equal to the number of of nonpivot columns of \(A\). Therefore, the sum is equal to the total number of columns of \(A\).
Theorem: If \(H\) is a \(p\)-dimensional subspace of \(\mathbb{R}^n\), then any linearly independent set of exactly \(p\) vectors in \(H\) is a basis of \(H\). We also have: any set of \(p\) vectors in \(H\) that spans \(H\) is a basis of \(H\).
Last modified on March 29, 2021
