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Chapter 6: Problem 12

Let \(A\) be a symmetric positive definite matrix. Show that the diagonal elements of \(A\) must all be positive.

Short Answer

Expert verified
In order to show that the diagonal elements of a symmetric positive definite matrix \(A\) must all be positive, we first recall that a matrix \(A\) is symmetric positive definite if the quadratic form \(x^TAx > 0\) for all non-zero vectors \(x\). We then use the standard basis vectors \(e_i\) and evaluate the quadratic form for them, obtaining \(e_i^T A e_i = a_{ii}\). By applying the positive definite condition, we have \(a_{ii} > 0\) for each standard basis vector \(e_i\). This implies that all diagonal elements of a symmetric positive definite matrix \(A\) must be positive.

Step by step solution

01

Definition of a symmetric positive definite matrix

Recall the definition of a symmetric positive definite matrix: A square matrix \(A\) is symmetric positive definite if it is symmetric, i.e., \(A^T = A\), and the quadratic form \(x^TAx > 0\) for all non-zero vectors \(x\).
02

Use standard basis vectors to obtain information about the diagonal elements

We want to extract information about the diagonal elements of \(A\), which are given by \(a_{ii}\). To do this, we can use the standard basis vectors \(e_i\), which have all zeros except for a 1 in the i-th entry. The quadratic form for standard basis vector \(e_i\) is given by: \[ e_i^T A e_i \]
03

Evaluate the quadratic form for standard basis vectors

Let's evaluate \(e_i^T A e_i\) for the given matrix \(A\). Since \(e_i\) is a vector with 1 at the i-th position and zeros elsewhere, multiplying \(A\) with \(e_i\) would just select the i-th column of \(A\), denoted as \(A_i\): \[ A e_i = A_i \] Now, multiply \(e_i^T\) with the result: \[ e_i^T A e_i = e_i^T A_i \] Once again, since \(e_i\) has 1 at the i-th position and zeros elsewhere, this multiplication just picks the i-th entry of \(A_i\), which is the diagonal element \(a_{ii}\): \[ e_i^T A e_i = a_{ii} \]
04

Apply the positive definite condition

Now, we can apply the positive definite condition on the result from Step 3. For every non-zero vector \(x\), including the standard basis vectors \(e_i\), it must hold that: \[ x^T A x > 0 \] Therefore, for each standard basis vector \(e_i\), since \(e_i^T A e_i = a_{ii}\): \[ a_{ii} > 0 \]
05

Conclusion

We have shown that, for a symmetric positive definite matrix \(A\), applying the positive definite condition on the quadratic form with standard basis vectors implies that all diagonal elements of \(A\) must be positive. Thus, if \(A\) is a symmetric positive definite matrix, its diagonal elements must all be positive.

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Most popular questions from this chapter

Let \(A\) be a symmetric positive definite matrix and let \(Q\) be an orthogonal diagonalizing matrix. Use the factorization \(A=Q D Q^{T}\) to find a nonsingular matrix \(B\) such that \(B^{T} B=A\)

Let \(A\) be an \(n \times n\) stochastic matrix and let e be the vector in \(\mathbb{R}^{n}\) whose entries are all equal to 1 Show that \(\mathbf{e}\) is an eigenvector of \(A^{T}\). Explain why a stochastic matrix must have \(\lambda=1\) as an eigenvalue.

Show that if \(A\) is symmetric positive definite, then \(\operatorname{det}(A)>0 .\) Give an example of a \(2 \times 2\) matrix with positive determinant that is not positive definite.

For each of the following, find a matrix \(B\) such that \(B^{2}=A\) (a) \(A=\left(\begin{array}{rr}2 & 1 \\ -2 & -1\end{array}\right)\) (b) \(A=\left(\begin{array}{rrr}9 & -5 & 3 \\ 0 & 4 & 3 \\ 0 & 0 & 1\end{array}\right)\)

For each of the following, find all possible values of the scalar \(\alpha\) that make the matrix defective or show that no such values exist: (a) \(\left(\begin{array}{lll}1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & \alpha\end{array}\right)\) (b) \(\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (c) \(\left(\begin{array}{rrr}1 & 2 & 0 \\ 2 & 1 & 0 \\ 2 & -1 & \alpha\end{array}\right)\) \((\mathbf{d})\left(\begin{array}{rrr}4 & 6 & -2 \\ -1 & -1 & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (e) \(\left(\begin{array}{ccc}3 \alpha & 1 & 0 \\ 0 & \alpha & 0 \\ 0 & 0 & \alpha\end{array}\right)\) (f) \(\left(\begin{array}{ccc}3 \alpha & 0 & 0 \\ 0 & \alpha & 1 \\ 0 & 0 & \alpha\end{array}\right)\) (g) \(\left(\begin{array}{ccc}\alpha+2 & 1 & 0 \\ 0 & \alpha+2 & 0 \\ 0 & 0 & 2 \alpha\end{array}\right)\) (h) \(\left(\begin{array}{ccc}\alpha+2 & 0 & 0 \\ 0 & \alpha+2 & 1 \\ 0 & 0 & 2 \alpha\end{array}\right)\)
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