Question

Consider a subspace $\mathit{V}$ in${\mathit{ℝ}}^{\mathbf{m}}$ that is defined by homogeneous linear equations

$\left|\begin{array}{l}{a}_{11}{x}_1+a_{12}{x}_2+\dots +a_{1m}{x}_m=0\\ a_{21}{x}_1+a_{22}{x}_2+\dots +a_{2m}{x}_m=0\\ ⋮⋮⋮⋮\\ a_{n1}{x}_1+a_{22}{x}_2+\dots +a_{\mathrm{nm}}{x}_m=0\end{array}\right|$.

What is the relationship between the dimension of $\mathit{V}$and the quantity$\mathit{m}\mathbf{-}\mathit{n}$

? State your answer as an inequality. Explain carefully.

Short answer

A subspace$V$ of ${ℝ}^m$has dimension at least $m-n$.

Step-by-step solution

To mention given data and recall the theorem 3.3.7

We have the subspace$V$ of${ℝ}^m$ defined by$n$ homogeneous linear eqautions,

role="math" localid="1660126682574" $\left|\begin{array}{l}{a}_{11}{x}_1+a_{12}{x}_2+\cdots +a_{1m}{x}_m=0\\ a_{21}{x}_1+a_{22}{x}_2+\cdots +a_{2m}{x}_m=0\\ ⋮⋮⋮⋮\\ a_{n1}{x}_1+a_{n2}{x}_2+\cdots +a_{nm}{x}_m=0\end{array}\right|$,

Therefore,$V$ can be written as

$V=\ker \left(A\right)$, where$A$ is an$n\times m$ matrix$A=\left[a_{ij}\right]$ .

Suppose the column vector be$\left[\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_m\end{array}\right]$ .

Theorem 3.3.7, is stated as follows:

For any$n\times m$ matrix $A$, the equation,

$\dim \left(\ker \left(\mathrm{A}\right)\right)+\dim \left(\mathrm{im}\left(\mathrm{A}\right)\right)=\mathrm{m}$.

To find the relationship between the dimension V and m-n 

We have the rank of $A$$rank\left(A\right)\le n$.

Thus, by Theorem 3.3.7, we have,

$\begin{array}{c}\dim \left(\mathrm{V}\right)=\dim \left(\ker \left(\mathrm{A}\right)\right)\\ =\mathrm{m}-\mathrm{rank}\left(\mathrm{A}\right)\\ \ge \mathrm{m}-\mathrm{n}\end{array}$

$\therefore \dim \left(\mathrm{V}\right)\ge \mathrm{m}-\mathrm{n}$

.

Hence, a subspace $V$of${\mathrm{ℝ}}^{\mathrm{m}}$ has dimension at least $m-n$.