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Chapter 6: Problem 31

Let \(A\) be a matrix whose columns all add up to a fixed constant \(\delta .\) Show that \(\delta\) is an eigenvalue of \(A\)

Short Answer

Expert verified
Let x be a column vector of length n with all entries equal to 1, i.e., x = \(\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}\). Then compute the matrix-vector product Ax, which is a vector with all entries as 未: Ax = \(\begin{bmatrix} \delta \\ \delta \\ \vdots \\ \delta \end{bmatrix}\). Since Ax = 未x, with x being a nonzero vector, 未 is an eigenvalue of A.

Step by step solution

01

Choose a vector x

Let x be a column vector of length n (where n is the number of rows in matrix A) with all entries equal to 1, i.e., x = \(\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}\).
02

Compute the product Ax

Now, calculate the matrix-vector product Ax. Since the columns of matrix A add up to 未, when we multiply A with the vector x, each entry of the resulting vector will be 未 (due to the dot product between the columns of A and the vector x). So, the product Ax is a vector with all entries as 未: Ax = \(\begin{bmatrix} \delta \\ \delta \\ \vdots \\ \delta \end{bmatrix}\).
03

Compute the product 未x

Multiply the scalar 未 by the vector x. Since all entries of x are 1, the product 未x is also a vector with all entries equal to 未: 未x = \(\begin{bmatrix} \delta \\ \delta \\ \vdots \\ \delta \end{bmatrix}\).
04

Comparing Ax and 未x

We've computed the products Ax and 未x and found that both are equal: Ax = 未x = \(\begin{bmatrix} \delta \\ \delta \\ \vdots \\ \delta \end{bmatrix}\). Since Ax = 未x and x is a nonzero vector, 未 is an eigenvalue of A according to the definition of an eigenvalue. Therefore, we have shown that fixed constant 未 is an eigenvalue of the matrix A.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix-Vector Multiplication
Matrix-vector multiplication is an operation where a matrix interacts with a vector, producing another vector. This process involves multiplying each column of the matrix by the respective elements of the vector and then summing these products into a single number for each row:
  • First, align the vector with the matrix columns.
  • Multiply each element in a matrix column by the corresponding vector element.
  • Sum these products across each row.
The resulting vector has as many elements as there are rows in the matrix. This operation is essential in understanding transformations in linear algebra.
Constant Vector
A constant vector is a vector where all its elements are the same value. In our exercise, the vector \(x\) is an example of a constant vector with every entry being 1:
  • This simplifies calculations since multiplying with other numbers doesn鈥檛 change the sameness of its entries.
  • It ensures the output follows a predictable pattern when used in matrix-vector multiplication, particularly with matrices that have specific structural properties.
Such vectors play a crucial role in verifying properties like eigenvalues, as seen in the exercise.
Matrix Columns
Matrix columns can be thought of as separate vectors within a matrix. Each column contributes to the matrix's properties, such as determining an eigenvalue. Here鈥檚 how each column plays a role:
  • When columns collectively add up to a constant like \( \delta \), it indicates a uniform influence across the entire matrix.
  • This uniformity is critical when assessing the product with a constant vector, revealing simplified results such as repeated entries in the resulting vector.
Understanding the makeup of these columns aids in explaining the behavior of a matrix during transformations.
Dot Product
The dot product is a fundamental operation in linear algebra that combines two vectors by multiplying their corresponding entries and summing the results. In matrix-vector multiplication:
  • The dot product of a matrix column with a vector determines each entry in the resulting vector.
  • It is crucial when calculating something like \(Ax\) in our exercise, since it helps convert the column sums of \(A\) into entries in the vector.
  • For a constant vector like \(x\), this multiplication process is straightforward as it repeatedly uses the same scalar, making each entry of \(Ax\) the column sum, equal to \( \delta \).
This operation consolidates the interaction between two vectors into a concise numerical result, crucial in many matrix operations.

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Most popular questions from this chapter

For each of the following pairs of vectors \(\mathbf{z}\) and \(\mathbf{w},\) compute (i) \(\|\mathbf{z}\|\) (ii) \(\|\mathbf{w}\|\) (iii) \(\langle\mathbf{z}, \mathbf{w}\rangle,\) and \((\text { iv })\langle\mathbf{w}, \mathbf{z}\rangle\) (a) \(\mathbf{z}=\left(\begin{array}{c}4+2 i \\ 4 i\end{array}\right), \quad \mathbf{w}=\left(\begin{array}{c}-2 \\ 2+i\end{array}\right)\) (b) \(\mathbf{z}=\left(\begin{array}{c}1+i \\ 2 i \\ 3-i\end{array}\right), \quad \mathbf{w}=\left(\begin{array}{c}2-4 i \\ 5 \\ 2 i\end{array}\right)\)

Prove each of the following: (a) If \(U\) is a unit upper triangular matrix, then \(U\) is nonsingular and \(U^{-1}\) is also unit upper triangular. (b) If \(U_{1}\) and \(U_{2}\) are both unit upper triangular matrices, then the product \(U_{1} U_{2}\) is also a unit upper triangular matrix.

Let \(A\) be an \(m \times n\) matrix of rank \(n\) with singular value decomposition \(U \Sigma V^{T}\). Let \(\Sigma^{+}\) denote the \(n \times m\) matrix \\[ \left[\begin{array}{ccccc} \frac{1}{\sigma_{1}} & & & \\ & \frac{1}{\sigma_{2}} & & & O \\ & & & \ddots & \\ & & & \frac{1}{\sigma_{n}} \end{array}\right] \\] and define \(A^{+}=V \Sigma^{+} U^{T} .\) Show that \(\hat{\mathbf{x}}=A^{+} \mathbf{b}\) satisfies the normal equations \(A^{T} A \mathbf{x}=A^{T} \mathbf{b}\)

Let \(A\) be a real \(2 \times 2\) matrix with the property that \(a_{21} a_{12}>0,\) and let \\[ r=\sqrt{a_{21} / a_{12}} \quad \text { and } \quad S=\left(\begin{array}{cc} r & 0 \\ 0 & 1 \end{array}\right) \\] Compute \(B=S A S^{-1} .\) What can you conclude about the eigenvalues and eigenvectors of \(B ?\) What can you conclude about the eigenvalues and eigenvectors of \(A\) ? Explain.

Given that \\[ A=\left(\begin{array}{ccc} 4 & 0 & 0 \\ 0 & 1 & i \\ 0 & -i & 1 \end{array}\right) \\] find a matrix \(B\) such that \(B^{H} B=A\)
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