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Chapter 4: Q4.5-27Q (page 191)

Explain why the space \[{\mathop{\rm P}\nolimits} \] of all polynomials is infinite-dimensional.

Short Answer

Expert verified

The space \[{\mathop{\rm P}\nolimits} \] of all polynomials is infinite-dimensional.

Step by step solution

01

State the condition for infinite dimension

If \(V\) is spanned by a finite set, it is said to be finite-dimensional. The dimension of V, written as \(\dim V\), is the number of vectors in abasis for \(V\). The dimension of the zero-vector space\(\left\{ 0 \right\}\)is defined to be zero. If \(V\) is not spanned by a finite set, it is said to beinfinite-dimensional.

02

Explain that \[{\mathop{\rm P}\nolimits} \] of all polynomials is an infinite-dimensional space

Consider \(\dim {\mathop{\rm P}\nolimits} = k < \infty \). Since \({{\mathop{\rm P}\nolimits} _n}\) is a subspace of \[{\mathop{\rm P}\nolimits} \] for all \(n\), and \(\dim k - 1 = k\), then \(\dim {{\mathop{\rm P}\nolimits} _{k - 1}} = \dim {\mathop{\rm P}\nolimits} \). This implies that \({{\mathop{\rm P}\nolimits} _{k - 1}} = {\mathop{\rm P}\nolimits} \), which is untrue. For example, \({\mathop{\rm p}\nolimits} \left( t \right) = {t^k}\) is in \[{\mathop{\rm P}\nolimits} \] but not in \[{{\mathop{\rm P}\nolimits} _{k - 1}}\]. Therefore, the dimension of \[{\mathop{\rm P}\nolimits} \] cannot be finite.

Thus, space \[{\mathop{\rm P}\nolimits} \] of all polynomials is infinite-dimensional.

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Most popular questions from this chapter

Let \({M_{2 \times 2}}\) be the vector space of all \(2 \times 2\) matrices, and define \(T:{M_{2 \times 2}} \to {M_{2 \times 2}}\) by \(T\left( A \right) = A + {A^T}\), where \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\).

  1. Show that \(T\)is a linear transformation.
  2. Let \(B\) be any element of \({M_{2 \times 2}}\) such that \({B^T} = B\). Find an \(A\) in \({M_{2 \times 2}}\) such that \(T\left( A \right) = B\).
  3. Show that the range of \(T\) is the set of \(B\) in \({M_{2 \times 2}}\) with the property that \({B^T} = B\).
  4. Describe the kernel of \(T\).

Find a basis for the set of vectors in\({\mathbb{R}^{\bf{3}}}\)in the plane\(x + {\bf{2}}y + z = {\bf{0}}\). (Hint:Think of the equation as a “system” of homogeneous equations.)

(M) Let \({{\mathop{\rm a}\nolimits} _1},...,{{\mathop{\rm a}\nolimits} _5}\) denote the columns of the matrix \(A\), where \(A = \left( {\begin{array}{*{20}{c}}5&1&2&2&0\\3&3&2&{ - 1}&{ - 12}\\8&4&4&{ - 5}&{12}\\2&1&1&0&{ - 2}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm a}\nolimits} _1}}&{{{\mathop{\rm a}\nolimits} _2}}&{{{\mathop{\rm a}\nolimits} _4}}\end{array}} \right)\)

  1. Explain why \({{\mathop{\rm a}\nolimits} _3}\) and \({{\mathop{\rm a}\nolimits} _5}\) are in the column space of \(B\).
  2. Find a set of vectors that spans \({\mathop{\rm Nul}\nolimits} A\).
  3. Let \(T:{\mathbb{R}^5} \to {\mathbb{R}^4}\) be defined by \(T\left( x \right) = A{\mathop{\rm x}\nolimits} \). Explain why \(T\) is neither one-to-one nor onto.

Let S be a maximal linearly independent subset of a vector space V. In other words, S has the property that if a vector not in S is adjoined to S, the new set will no longer be linearly independent. Prove that S must be a basis of V. [Hint: What if S were linearly independent but not a basis of V?]

Exercises 23-26 concern a vector space V, a basis \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)and the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\).

Given vectors, \({u_{\bf{1}}}\),….,\({u_p}\) and w in V, show that w is a linear combination of \({u_{\bf{1}}}\),….,\({u_p}\) if and only if \({\left( w \right)_B}\) is a linear combination of vectors \({\left( {{{\bf{u}}_{\bf{1}}}} \right)_B}\),….,\({\left( {{{\bf{u}}_p}} \right)_B}\).
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