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Let \(H\) be an \(n\)-dimensional subspace of an \(n\)-dimensional vector space \(V\). Show that \(H = V\).

Short Answer

Expert verified

It is proved that \(H = V\).

Step by step solution

01

State the basis theorem

Let \(V\) be a p-dimensional vector space, such that \(p \ge 1\). Anylinearly independent setof exactly \(p\) elements in \(V\) is automatically a basis for \(V\). Any set of exactly \(p\) elements that spans \(V\) is automatically a basis for \(V\).

02

Show that \(H = V\)

When \(\dim V = \dim H = 0\), \(V = \left\{ 0 \right\}\) and \(H = \left\{ 0 \right\}\), and thus, \(H = V\).

Consider \(\dim V = \dim H > 0\). Then, \(H\) contains a basis \(S\) of \(n\) vectors. However, \(S\) is also a basis for \(V\) according to the basis theorem. Hence, \(H = V = {\mathop{\rm Span}\nolimits} S\).

Thus, it is proved that \(H = V\).

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Most popular questions from this chapter

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 \({{\mathop{\rm x}\nolimits} _4} = {\mathop{\rm v}\nolimits} \).

If a\({\bf{6}} \times {\bf{3}}\)matrix A has a rank 3, find dim Nul A, dim Row A, and rank\({A^T}\).

(M) Let \(H = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) and \(K = {\mathop{\rm Span}\nolimits} \left\{ {{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\), where

\({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}5\\3\\8\end{array}} \right),{{\mathop{\rm v}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}1\\3\\4\end{array}} \right),{{\mathop{\rm v}\nolimits} _3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\5\end{array}} \right),{{\mathop{\rm v}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\{ - 12}\\{ - 28}\end{array}} \right)\)

Then \(H\) and \(K\) are subspaces of \({\mathbb{R}^3}\). In fact, \(H\) and \(K\) are planes in \({\mathbb{R}^3}\) through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line. (Hint: w can be written as \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2}\) and also as \({c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\). To build w, solve the equation \({c_1}{{\mathop{\rm v}\nolimits} _1} + {c_2}{{\mathop{\rm v}\nolimits} _2} = {c_3}{{\mathop{\rm v}\nolimits} _3} + {c_4}{{\mathop{\rm v}\nolimits} _4}\) for the unknown \({c_j}'{\mathop{\rm s}\nolimits} \).)

Let \(A\) be an \(m \times n\) matrix of rank \(r > 0\) and let \(U\) be an echelon form of \(A\). Explain why there exists an invertible matrix \(E\) such that \(A = EU\), and use this factorization to write \(A\) as the sum of \(r\) rank 1 matrices. [Hint: See Theorem 10 in Section 2.4.]

Prove theorem 3 as follows: Given an \(m \times n\) matrix A, an element in \({\mathop{\rm Col}\nolimits} A\) has the form \(Ax\) for some x in \({\mathbb{R}^n}\). Let \(Ax\) and \(A{\mathop{\rm w}\nolimits} \) represent any two vectors in \({\mathop{\rm Col}\nolimits} A\).

  1. Explain why the zero vector is in \({\mathop{\rm Col}\nolimits} A\).
  2. Show that the vector \(A{\mathop{\rm x}\nolimits} + A{\mathop{\rm w}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} A\).
  3. Given a scalar \(c\), show that \(c\left( {A{\mathop{\rm x}\nolimits} } \right)\) is in \({\mathop{\rm Col}\nolimits} A\).
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