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Chapter 1: Problem 13

Test to see if \(A^{\mathrm{T}} A\) is positive definite in each case: \(A\) needs independent columns. $$ A=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right] \quad \text { and } \quad A=\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \\ 2 & 1 \end{array}\right] \quad \text { and } \quad A=\left[\begin{array}{lll} 1 & 1 & 2 \\ 1 & 2 & 1 \end{array}\right] $$

Short Answer

Expert verified
The first two matrices result in positive definite matrices, while the third one does not.

Step by step solution

01

Transpose each matrix A

First, find the transpose of each given matrix \( A \). If \( A = \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix} \), then the transpose \( A^{\mathrm{T}} = \begin{bmatrix} 1 & 0 \ 2 & 3 \end{bmatrix} \). For \( A = \begin{bmatrix} 1 & 1 \ 1 & 2 \ 2 & 1 \end{bmatrix} \), the transpose is \( A^{\mathrm{T}} = \begin{bmatrix} 1 & 1 & 2 \ 1 & 2 & 1 \end{bmatrix} \). For \( A = \begin{bmatrix} 1 & 1 & 2 \ 1 & 2 & 1 \end{bmatrix} \), the transpose is \( A^{\mathrm{T}} = \begin{bmatrix} 1 & 1 \ 1 & 2 \ 2 & 1 \end{bmatrix} \).
02

Compute the matrix product \(A^{\mathrm{T}}A\)

Multiply \( A^{\mathrm{T}} \) with \( A \) for each transposed matrix to calculate \( A^{\mathrm{T}} A \). For the first matrix, \( A^{\mathrm{T}} A = \begin{bmatrix} 1 & 0 \ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 2 & 13 \end{bmatrix} \).For the second matrix, \( A^{\mathrm{T}} A = \begin{bmatrix} 1 & 1 & 2 \ 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 1 & 2 \ 2 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 5 \ 5 & 6 \end{bmatrix} \). For the third matrix, \( A^{\mathrm{T}} A = \begin{bmatrix} 1 & 1 \ 1 & 2 \ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 2 \ 1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 3 \ 3 & 5 & 4 \ 3 & 4 & 5 \end{bmatrix} \).
03

Determine the positive definiteness of \(A^{\mathrm{T}}A\)

A matrix is positive definite if all of its eigenvalues are positive, or by Sylvester's criterion, all its leading principal minors have positive determinants.For the first matrix product, \((A^{\mathrm{T}} A) = \begin{bmatrix} 1 & 2 \ 2 & 13 \end{bmatrix} \), calculate the determinants of leading principal minors: - The first leading principal minor \(1 > 0\).- The determinant \( \det(\begin{bmatrix} 1 & 2 \ 2 & 13 \end{bmatrix}) = 1 \times 13 - 2 \times 2 = 9 > 0\).Thus, \(A^{\mathrm{T}}A\) is positive definite.For the second matrix product, \((A^{\mathrm{T}} A) = \begin{bmatrix} 6 & 5 \ 5 & 6 \end{bmatrix} \), calculate:- The first leading principal minor \(6 > 0\).- The determinant \( \det(\begin{bmatrix} 6 & 5 \ 5 & 6 \end{bmatrix}) = 6 \times 6 - 5 \times 5 = 11 > 0\).Thus, \(A^{\mathrm{T}}A\) is positive definite.For the third matrix product, \((A^{\mathrm{T}} A) = \begin{bmatrix} 2 & 3 & 3 \ 3 & 5 & 4 \ 3 & 4 & 5 \end{bmatrix} \), check:- The first leading principal minor \(2 > 0\).- The determinant of the first two steps principal minor \( \begin{bmatrix} 2 & 3 \ 3 & 5 \end{bmatrix} = 2 \times 5 - 3 \times 3 = 1 > 0\).- However, compute the determinant of the full 3x3 matrix or eigenvalues to confirm positivity.By examining, the determinant is zero indicating \(A^{\mathrm{T}}A\) is not positive definite.

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Key Concepts

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

Independent Columns
In linear algebra, matrix columns are called independent if none of the columns can be expressed as a linear combination of the others. This concept is crucial when determining if a matrix has full rank, meaning all its columns are necessary and introduce new information. When you have independent columns in matrix \(A\), it implies the matrix has maximal column rank. Having independent columns is essential for \(A^{\mathrm{T}}A\) (transpose of A times A) to be positive definite. Positive definiteness implies several properties that are vital for matrix stability and invertibility. To test for column independence, you can use methods like:
  • Row reduction techniques to identify pivot positions.
  • Checking if the determinant (for square matrices) is non-zero.
When dealing with square matrices, full rank ensures a non-zero determinant, supporting positive definiteness.
Transpose of a Matrix
The transpose operation is fundamental in linear algebra, written as \(A^{\mathrm{T}}\). Transposing a matrix involves flipping it over its diagonal. This means rows become columns and vice versa. For example, if\[A = \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix}, \]then its transpose is\[A^{\mathrm{T}} = \begin{bmatrix} 1 & 0 \ 2 & 3 \end{bmatrix}.\]Transposing is useful for various calculations, including testing a matrix for symmetry, and it plays a key role when determining the positive definiteness of \(A^{\mathrm{T}}A\). The importance of the transpose becomes evident when dealing with vector spaces and proofs in eigenvalues and eigenvectors. Symmetry in \(A^{\mathrm{T}}A\) is particularly handy because symmetry is a required property for a matrix to be positive definite. It's easy to remember: "transposing" means "flipping" across the diagonal.
Eigenvalues
Eigenvalues are a central concept in linear algebra. They arise from the equation \(A\mathbf{v} = \lambda\mathbf{v}\), where \(A\) is a matrix, \(\lambda\) represents eigenvalues, and \(\mathbf{v}\) (non-zero) is an eigenvector. Essentially, they indicate the factor by which the eigenvector is stretched or shrunk when transformed by \(A\).For a matrix to be positive definite, all its eigenvalues should be positive. If even one of these eigenvalues is zero or negative, the matrix cannot be deemed positive definite. Positive eigenvalues ensure that all directions in the space spanned by the matrix have positive curvature, which is a requirement for stability and invertibility.You typically find eigenvalues through characteristic equations, commonly solved by finding roots of the determinant \[\det(A - \lambda I) = 0.\]Positive definiteness via eigenvalues ensures that transformations involving \(A\) do not flip or flatten dimensions of your data.
Sylvester's Criterion
Sylvester's Criterion provides a practical way to determine if a matrix is positive definite. It states that a matrix is positive definite if and only if all its leading principal minors 鈥� the determinants of the top-left submatrices 鈥� are positive. This method is applied step-by-step by examining:
  • The first leading principal minor, which is simply the top-left element of the matrix.
  • Subsequent principal minors formed by slightly larger submatrices of the matrix.
For example, for a 2x2 matrix \( \begin{bmatrix} a & b \ b & c \end{bmatrix} \), this involves checking that \(a > 0\) and \(\det(\begin{bmatrix} a & b \ b & c \end{bmatrix}) > 0\).Using Sylvester鈥檚 Criterion is effective as it removes the need to directly compute eigenvalues. By recalculating determinants iteratively, you get a manageable check for definiteness. It not only confirms positivity for quadratic forms but also brings ease in matrix computations across various fields, including statistics and machine learning.

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Most popular questions from this chapter

If \(a_{11}, \ldots, a_{1 n}\) is the first row of a rank-1 matrix \(A\) and \(a_{11}, \ldots, a_{m 1}\) is the first column, find a formula for \(a_{i j}\). Good to check when \(a_{11}=2, a_{12}=3, a_{21}=4\). When will your formula break down? Then rank 1 is impossible or not unique.

The second column of \(A-U V\) is \(\left[\begin{array}{l}c \\\ d\end{array}\right]-\left[\begin{array}{l}u_{1} \\ u_{2}\end{array}\right] v_{2}=a_{2}-v_{2} u\). Which number \(v_{2}\) minimizes \(\left\|a_{2}-v_{2} u\right\|^{2}\) ? $$ \text { Vector form } \quad \text { The best } \boldsymbol{v}=\left[\begin{array}{ll} v_{1} & v_{2} \end{array}\right] \text { solves }\left(\boldsymbol{u}^{\mathrm{T}} \boldsymbol{u}\right) \boldsymbol{v}=\boldsymbol{u}^{\mathrm{T}} A $$

(Important) Show that \(A^{\mathrm{T}} A\) has the same nullspace as \(A\). Here is one approach: First, if \(A x\) equals zero then \(A^{\mathrm{T}} A x\) equals This proves \(\mathbf{N}(A) \subset \mathbf{N}\left(A^{\mathrm{T}} A\right)\). Second, if \(A^{\mathrm{T}} A x=\mathbf{0}\) then \(\boldsymbol{x}^{\mathrm{T}} A^{\mathrm{T}} A \boldsymbol{x}=\|A \boldsymbol{x}\|^{2}=0 .\) Deduce \(\mathbf{N}\left(A^{\mathrm{T}} A\right)=\mathbf{N}(A) .\)

In the Cholesky factorization \(S=A^{\mathrm{T}} A\), with \(A=\sqrt{D} L^{\mathrm{T}}\), the square roots of the pivots are on the diagonal of \(A\). Find \(A\) (upper triangular) for $$ S=\left[\begin{array}{lll} 9 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 8 \end{array}\right] \quad \text { and } \quad S=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 7 \end{array}\right] $$

The eigenvalues of \(A\) equal the eigenvalues of \(A^{T}\). This is because \(\operatorname{det}(A-\lambda I)\) equals \(\operatorname{det}\left(A^{T}-A I\right)\). That is true because Show by an example that the eigenvectors of \(A\) and \(A^{\mathrm{T}}\) are not the same.
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