Collection of Maths Problems

Operations on positive definite matrices

Task number: 2729

• Variant 1

  The matrix \(\mathbf A+\mathbf B\) is also positive definite.

• Resolution

  \(\mathbf x^H(\mathbf A+\mathbf B)\mathbf x=\mathbf x^H \mathbf A\mathbf x + \mathbf x^H\mathbf B \mathbf x> 0\) for \(\mathbf x\ne \mathbf 0\).

• Variant 2

  The matrix \(\mathbf A^{-1}\) is also positive definite (i.e. show also that \(\mathbf A\) is regular).

• Resolution

  The system \(\mathbf A\mathbf x=\mathbf 0\) has only trivial solution, i.e. the matrix \(\mathbf A\) is regular, and \(\mathbf A^{-1}\) exists.

  Then for \(\mathbf x\ne\mathbf 0\) we get
  \(\mathbf x^H\mathbf A^{-1}\mathbf x=\mathbf x^H\mathbf A^{-1}\mathbf A\mathbf A^{-1}\mathbf x=\mathbf y^H\mathbf A\mathbf y>0\) for \(\mathbf y=\mathbf A^{-1}\mathbf x\ne\mathbf 0\).

  Observe that for \(\mathbf y^H\) we have used the fact that
  \((\mathbf A^{-1})^H=(\mathbf A^{-1})^H(\mathbf A\mathbf A^{-1})=((\mathbf A^{-1})^H\mathbf A^H)\mathbf A^{-1}=(\mathbf A\mathbf A^{-1})^H\mathbf A^{-1}=\mathbf A^{-1}\).

  Alternatively, if \(\mathbf A=\mathbf U^H\mathbf U\) is a product of regular matrices, is
  \(\mathbf A^{-1}=(\mathbf U^H\mathbf U)^{-1}=\mathbf U^{-1}(\mathbf U^H)^{-1}=\mathbf U^{-1}(\mathbf U^{-1})^H\),
  that is a product of regular matrices \(\mathbf A^{-1}\) is also positive definite.

  We may also use the similarity argument. For a positive definite \(\mathbf A\) holds that \(\mathbf A=\mathbf R^H \mathbf D \mathbf R\) for unitary \(\mathbf R\). Then \(\mathbf A^{-1}=\mathbf R^H \mathbf D^{-1}\mathbf R\) is also regular, hermitian and positive definite.

  Ad product: the product of two positive definite matrices need not to be positive definite, it need not to be symmetric, indeed:

  \[ \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 4 \\ \end{pmatrix} \cdot \begin{pmatrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \\ \end{pmatrix} = \begin{pmatrix} 7 & 5 & 1 \\ 5 & 7 & 2 \\ 0 & 4 & 12 \\ \end{pmatrix} \]

  If the product was symmetric (Hermitian), then it would be positive definite. (It follows from the simultaneous diagonalization of normal matrices, which is beyond the scope of the course LA II.)