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Suppose \({\mathbb{R}^4} = Span\left\{ {{v_1},...,{v_4}} \right\}\). Explain why \(\left\{ {{v_1},...,{v_4}} \right\}\) is a basis for \({\mathbb{R}^4}\).

Short Answer

Expert verified

The set \(\left\{ {{v_1},...,{v_4}} \right\}\) is a basis for \({\mathbb{R}^4}\).

Step by step solution

01

State the invertible matrix theorem

Recall the invertible matrix theorem which states that if the square matrix is invertible, then the columns are linearly independent, and the columns form a basis for \({\mathbb{R}^n}\).

02

State the basis theorem

It is given that\(\left\{ {{v_1},...,{v_4}} \right\}\)is the set of vectors. It forms a square matrix of the order\(4 \times 4\)as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}}&{{{\bf{v}}_4}}\end{array}} \right]\)

As\({\mathbb{R}^4} = {\rm{Span}}\left\{ {{v_1},...,{v_4}} \right\}\), the columns of A span\({\mathbb{R}^4}\). Also, the columns are linearly independent and form a basis.

Thus, \(\left\{ {{v_1},...,{v_4}} \right\}\) is a basis for \({\mathbb{R}^4}\).

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Most popular questions from this chapter

Suppose a nonhomogeneous system of nine linear equations in ten unknowns has a solution for all possible constants on the right sides of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Discuss.

Question 18: Suppose A is a \(4 \times 4\) matrix and B is a \(4 \times 2\) matrix, and let \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) represent a sequence of input vectors in \({\mathbb{R}^2}\).

  1. Set \({{\mathop{\rm x}\nolimits} _0} = 0\), compute \({{\mathop{\rm x}\nolimits} _1},...,{{\mathop{\rm x}\nolimits} _4}\) from equation (1), and write a formula for \({{\mathop{\rm x}\nolimits} _4}\) involving the controllability matrix \(M\) appearing in equation (2). (Note: The matrix \(M\) is constructed as a partitioned matrix. Its overall size here is \(4 \times 8\).)
  2. Suppose \(\left( {A,B} \right)\) is controllable and v is any vector in \({\mathbb{R}^4}\). Explain why there exists a control sequence \({{\mathop{\rm u}\nolimits} _0},...,{{\mathop{\rm u}\nolimits} _3}\) in \({\mathbb{R}^2}\) such that \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

16. If \(A\) is an \(m \times n\) matrix of rank\(r\), then a rank factorization of \(A\) is an equation of the form \(A = CR\), where \(C\) is an \(m \times r\) matrix of rank\(r\) and \(R\) is an \(r \times n\) matrix of rank \(r\). Such a factorization always exists (Exercise 38 in Section 4.6). Given any two \(m \times n\) matrices \(A\) and \(B\), use rank factorizations of \(A\) and \(B\) to prove that rank\(\left( {A + B} \right) \le {\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B\).

(Hint: Write \(A + B\) as the product of two partitioned matrices.)

(M) Determine whether w is in the column space of \(A\), the null space of \(A\), or both, where

\({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 3}\end{array}} \right),A = \left( {\begin{array}{*{20}{c}}7&6&{ - 4}&1\\{ - 5}&{ - 1}&0&{ - 2}\\9&{ - 11}&7&{ - 3}\\{19}&{ - 9}&7&1\end{array}} \right)\)

The null space of a \({\bf{5}} \times {\bf{6}}\) matrix A is 4-dimensional, what is the dimension of the column space of A.

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