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Suppose a \({\bf{4}} \times {\bf{7}}\) matrix A has three pivot columns. Is Col \(A = {\mathbb{R}^{\bf{3}}}\)? Is Nul \(A = {\mathbb{R}^{\bf{2}}}\)? Explain your answers.

Short Answer

Expert verified

The dimension of A is 4.

Step by step solution

01

Find the basis of Col A and Nul A

Since the pivot column of A forms a basis of Col A, it suggests that three pivot columns form a basis of Col A.

\({\rm{Col}}\;A = {\mathbb{R}^3}\)

02

Use the rank theorem to find the dimensions of Nul A

Since the pivot columns of A form a basis of Col A, the dimensions of Nul A can be obtained using the rank theorem.

\(\begin{array}{c}{\rm{rank}}\;A + \dim \left( {{\rm{Nul}}\,A} \right) = n\\3 + \dim \left( {{\rm{Nul}}\;A} \right) = 7\\\dim \left( {{\rm{Nul}}\;A} \right) = 4\end{array}\)

So, the dimension of A is 4.

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