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Chapter 2: Q11SE (page 93)

Let \({x_1},...,{x_n}\) be fixed numbers. The matrix below called a Vandermonde matrix, occurs in applications such as signal processing, error-correcting codes, and polynomial interpolation.

\(V = \left( {\begin{aligned}{*{20}{c}}1&{{x_1}}&{x_1^2}& \ldots &{x_1^{n - 1}}\\1&{{x_2}}&{x_2^2}& \ldots &{x_2^{n - 1}}\\ \vdots & \vdots & \vdots &{}& \vdots \\1&{{x_n}}&{x_n^2}& \ldots &{x_n^{n - 1}}\end{aligned}} \right)\)

Given \(y = \left( {{y_1},...,{y_n}} \right)\) in \({\mathbb{R}^n}\), suppose \({\mathop{\rm c}\nolimits} = \left( {{c_0},...,{c_{n - 1}}} \right)\) in \({\mathbb{R}^n}\) satisfies \(V{\mathop{\rm c}\nolimits} = {\mathop{\rm y}\nolimits} \), and define the polynomial.

\(p\left( t \right) = {c_0} + {c_1}t + {c_2}{t^2} + .... + {c_{n - 1}}{t^{n - 1}}\).

a. Show that \(p\left( {{x_1}} \right) = {{\mathop{\rm y}\nolimits} _1},...,p\left( {{x_n}} \right) = {{\mathop{\rm y}\nolimits} _n}\). We call \(p\left( t \right)\) an interpolating polynomial for the points \(\left( {{x_1},{y_1}} \right),...,\left( {{x_n},{y_n}} \right)\) because the graph of \(p\left( t \right)\) passes through the points.

b. Suppose \({x_1},...,{x_n}\) are distinct numbers. Show that the columns of V are linearly independent. (Hint: How many zeros can a polynomial of degree \(n - 1\) have?)

c. Prove: 鈥淚f \({x_1},...,{x_n}\) are distinct numbers, and \({y_1},...,{y_n}\) are arbitrary numbers, then there is an interpolating polynomial of degree \( \le n - 1\) for \(\left( {{x_1},{y_1}} \right),...,\left( {{x_n},{y_n}} \right)\).鈥�

Short Answer

Expert verified
  1. It is proved that \(p\left( {{x_1}} \right) = {{\mathop{\rm y}\nolimits} _1},...,p\left( {{x_n}} \right) = {{\mathop{\rm y}\nolimits} _n}\).
  2. It is proved that the columns of V are linearly independent.
  3. It is proved that \(p\) is an interpolating polynomial for \(\left( {{x_1},{y_1}} \right),...,\left( {{x_n},{y_n}} \right)\).

Step by step solution

01

Show that \(p\left( {{x_1}} \right) = {{\mathop{\rm y}\nolimits} _1},...,p\left( {{x_n}} \right) = {{\mathop{\rm y}\nolimits} _n}\)(a)

It is given that \(y = \left( {{y_1},...,{y_n}} \right)\) in \({\mathbb{R}^n}\). Suppose \({\mathop{\rm c}\nolimits} = \left( {{c_0},...,{c_{n - 1}}} \right)\)in \({\mathbb{R}^n}\) satisfies \(V{\mathop{\rm c}\nolimits} = {\mathop{\rm y}\nolimits} \).

The case of \(i = 1,..,n\)is shown below.

\(\begin{aligned}{c}p\left( {{x_i}} \right) = {c_0} + {c_1}{x_i} + ... + {c_{n - 1}}x_i^{n - 1}\\ = {{\mathop{\rm row}\nolimits} _i}\left( V \right) \cdot \left( {\begin{aligned}{*{20}{c}}{{c_0}}\\ \vdots \\{{c_{n - 1}}}\end{aligned}} \right)\\ = {{\mathop{\rm row}\nolimits} _i}\left( V \right){\mathop{\rm c}\nolimits} \end{aligned}\)

According to the property of matrix multiplication, the fact that c was chosen to satisfy \(V{\mathop{\rm c}\nolimits} = {\mathop{\rm y}\nolimits} \)is shown below.

\(\begin{aligned}{c}{{\mathop{\rm row}\nolimits} _i}\left( V \right){\mathop{\rm c}\nolimits} = {{\mathop{\rm row}\nolimits} _i}\left( {V{\mathop{\rm c}\nolimits} } \right)\\ = {{\mathop{\rm row}\nolimits} _i}\left( y \right)\\ = {y_i}\end{aligned}\)

Therefore\(p\left( {{x_i}} \right) = {y_i}\). To conclude, the entries in \(V{\mathop{\rm c}\nolimits} \) represent the values of the polynomial \(p\left( x \right)\) at \({x_1},...,{x_n}\).

Hence, it is proved that \(p\left( {{x_1}} \right) = {{\mathop{\rm y}\nolimits} _1},...,p\left( {{x_n}} \right) = {{\mathop{\rm y}\nolimits} _n}\).

02

Show that the columns of V are linearly independent(b)

Let \({x_1},...,{x_n}\) be distinct vectors and let \(Vc = 0\) for some vector c. Therefore, the entries in c represent the polynomial whose coefficients are zero at the distinct points\({x_1},...,{x_n}\). In addition, the polynomial must be identically zero since a non-zero polynomial of degree \(n - 1\) cannot have \(n\) zeros. This means that the entries in \(c\) must all be zero. It demonstrates that the columns of V are linearly independent.

Thus, it is proved that the columns of V are linearly independent.

03

Show that the columns of V are linearly independent(c)

If \({x_1},...,{x_n}\) are distinct vectors, then the columns of V are linearly independent by part (b). V is invertible, and the columns of V span \({\mathbb{R}^n}\) according to the invertible matrix theorem. Therefore, for every \(y = \left( {{y_1},...,{y_n}} \right)\) in \({\mathbb{R}^n}\), there exists a vector c such that \(Vc = y\). Suppose p be the polynomial whose coefficients are contained in c. According to part (a), \(p\) is an interpolating polynomial for \(\left( {{x_1},{y_1}} \right),...,\left( {{x_n},{y_n}} \right)\).

Thus, it is proved that \(p\) is an interpolating polynomial for \(\left( {{x_1},{y_1}} \right),...,\left( {{x_n},{y_n}} \right)\).

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Most popular questions from this chapter

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\)as\(n \times 1\)matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is a \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

28. If u and v are in \({\mathbb{R}^n}\), how are \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) related? How are \({{\mathop{\rm uv}\nolimits} ^T}\) and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) related?

Let Abe an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If Ais larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5鈥�8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

7. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\{\bf{0}}&{\bf{0}}\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.
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