8.3E: Positive Definite Matrices Exercises

Exercises for 1

solutions

2

Find the Cholesky decomposition of each of the following matrices.

\(\left[ \begin{array}{rr} 4 & 3 \\ 3 & 5 \end{array}\right]\) \(\left[ \begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right]\) \(\left[ \begin{array}{rrr} 12 & 4 & 3 \\ 4 & 2 & -1 \\ 3 & -1 & 7 \end{array}\right]\) \(\left[ \begin{array}{rrr} 20 & 4 & 5 \\ 4 & 2 & 3 \\ 5 & 3 & 5 \end{array}\right]\)

2. \(U = \frac{\sqrt{2}}{2} \left[ \begin{array}{rr} 2 & -1 \\ 0 & 1 \end{array}\right]\)
3. \(U = \frac{1}{30} \left[ \begin{array}{ccc} 60\sqrt{5} & 12\sqrt{5} & 15\sqrt{5} \\ 0 & 6\sqrt{30} & 10\sqrt{30} \\ 0 & 0 & 5\sqrt{15} \end{array}\right]\)

1. If \(A\) is positive definite, show that \(A^{k}\) is positive definite for all \(k \geq 1\).
2. Prove the converse to (a) when \(k\) is odd.
3. Find a symmetric matrix \(A\) such that \(A^{2}\) is positive definite but \(A\) is not.

2. If \(\lambda^{k} > 0\), \(k\) odd, then \(\lambda > 0\).

Let \(A = \left[ \begin{array}{rr} 1 & a \\ a & b \end{array}\right]\). If \(a^{2} < b\), show that \(A\) is positive definite and find the Cholesky factorization.

If \(A\) and \(B\) are positive definite and \(r > 0\), show that \(A + B\) and \(rA\) are both positive definite.

If \(\mathbf{x} \neq \mathbf{0}\), then \(\mathbf{x}^{T}A\mathbf{x} > 0\) and \(\mathbf{x}^{T}B\mathbf{x} > 0\). Hence \(\mathbf{x}^{T}(A + B)\mathbf{x} = \mathbf{x}^{T}A\mathbf{x} + \mathbf{x}^{T}B\mathbf{x} > 0\) and \(\mathbf{x}^{T}(rA)\mathbf{x} = r(\mathbf{x}^{T}A\mathbf{x}) > 0\), as \(r > 0\).

If \(A\) and \(B\) are positive definite, show that \(\left[ \begin{array}{rr} A & 0 \\ 0 & B \end{array}\right]\) is positive definite.

If \(A\) is an \(n \times n\) positive definite matrix and \(U\) is an \(n \times m\) matrix of rank \(m\), show that \(U^{T}AU\) is positive definite.

Let \(\mathbf{x} \neq \mathbf{0}\) in \(\mathbb{R}^n\). Then \(\mathbf{x}^{T}(U^{T}AU)\mathbf{x} = (U\mathbf{x})^{T}A(U\mathbf{x}) > 0\) provided \(U\mathbf{x} \neq 0\). But if \(U = \left[ \begin{array}{cccc} \mathbf{c}_{1} & \mathbf{c}_{2} & \dots & \mathbf{c}_{n} \end{array}\right]\) and \(\mathbf{x} = (x_{1}, x_{2}, \dots, x_{n})\), then \(U\mathbf{x} = x_{1}\mathbf{c}_{1} + x_{2}\mathbf{c}_{2} + \dots + x_{n}\mathbf{c}_{n} \neq \mathbf{0}\) because \(\mathbf{x} \neq \mathbf{0}\) and the \(\mathbf{c}_{i}\) are independent.

If \(A\) is positive definite, show that each diagonal entry is positive.

Let \(A_{0}\) be formed from \(A\) by deleting rows 2 and 4 and deleting columns 2 and 4. If \(A\) is positive definite, show that \(A_{0}\) is positive definite.

If \(A\) is positive definite, show that
\(A = CC^{T}\) where \(C\) has orthogonal columns.

If \(A\) is positive definite, show that \(A = C^{2}\) where \(C\) is positive definite.

Let \(P^{T}AP = D = \func{diag}(\lambda_{1}, \dots, \lambda_{n})\) where \(P^{T} = P\). Since \(A\) is positive definite, each eigenvalue \(\lambda_{i} > 0\). If \(B = \func{diag}(\sqrt{\lambda_{1}}, \dots, \sqrt{\lambda_{n}})\) then \(B^{2} = D\), so \(A = PB^{2}P^{T} = (PBP^{T})^{2}\). Take \(C = PBP^{T}\). Since \(C\) has eigenvalues \(\sqrt{\lambda_{i}} > 0\), it is positive definite.

Let \(A\) be a positive definite matrix. If \(a\) is a real number, show that \(aA\) is positive definite if and only if \(a > 0\).

1. Suppose an invertible matrix \(A\) can be factored in \(\mathbf{M}_{nn}\) as \(A = LDU\) where \(L\) is lower triangular with \(1\)s on the diagonal, \(U\) is upper triangular with \(1\)s on the diagonal, and \(D\) is diagonal with positive diagonal entries. Show that the factorization is unique: If \(A = L_{1}D_{1}U_{1}\) is another such factorization, show that \(L_{1} = L\), \(D_{1} = D\), and \(U_{1} = U\).
2. Show that a matrix \(A\) is positive definite if and only if \(A\) is symmetric and admits a factorization \(A = LDU\) as in (a).

2. If \(A\) is positive definite, use Theorem [thm:024815] to write \(A = U^{T}U\) where \(U\) is upper triangular with positive diagonal \(D\). Then \(A = (D^{-1}U)^{T}D^{2}(D^{-1}U)\) so \(A = L_{1}D_{1}U_{1}\) is such a factorization if \(U_{1} = D^{-1}U\), \(D_{1} = D^{2}\), and \(L_{1} = U^T_1\). Conversely, let \(A^{T} = A = LDU\) be such a factorization. Then \(U^{T}D^{T}L^{T} = A^{T} = A = LDU\), so \(L = U^{T}\) by (a). Hence \(A = LDL^{T} = V^{T}V\) where \(V = LD_{0}\) and \(D_{0}\) is diagonal with \(D^2_0 = D\) (the matrix \(D_{0}\) exists because \(D\) has positive diagonal entries). Hence \(A\) is symmetric, and it is positive definite by Example [exa:024865].

Let \(A\) be positive definite and write \(d_{r} = \det {^{(r)}A}\) for each \(r = 1, 2, \dots, n\). If \(U\) is the upper triangular matrix obtained in step 1 of the algorithm, show that the diagonal elements \(u_{11}, u_{22}, \dots, u_{nn}\) of \(U\) are given by \(u_{11} = d_{1}\), \(u_{jj} = d_{j} / d_{j-1}\) if \(j > 1\). [Hint: If \(LA = U\) where \(L\) is lower triangular with \(1\)s on the diagonal, use block multiplication to show that \(\det {^{(r)}A} = \det {^{(r)}U}\) for each \(r\).]