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Q8E
Expert-verifiedSuppose a \({\bf{5}} \times {\bf{6}}\) matrix A has four pivot columns. What is dim Nul A? Is \({\bf{Col}}\,A = {\mathbb{R}^{\bf{3}}}\)? Why or why not?
dim Nul A=2, \({\rm{Col}}\,A \ne {\mathbb{R}^4}\)
A has four pivot columns. By using the rank theorem, you get:
\(\begin{aligned}{c}{\rm{rank}}\,A + \dim \,{\rm{Nul}}\,A &= 6\\4 + \dim \,{\rm{Nul}}\,A &= 6\\\dim \,{\rm{Nul}}\,A &= 2\end{aligned}\)
The dim Col A =4, but Col A is a subspace of \({\mathbb{R}^5}\). Therefore, \({\rm{Col}}\,A \ne {\mathbb{R}^4}\),
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