Characterisation of Real Symmetric Positive Definite Matrix
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Theorem
Let $A$ be an $n \times n$ symmetric matrix over $\mathbb R$.
Then $A$ is positive definite if and only if:
- there exists a nonsingular matrix $C$ such that $A = C^\intercal C$.
Proof
Necessary Condition
Let $A$ be positive definite.
From Real Symmetric Matrix is Orthogonally Diagonalizable:
- there exists an orthogonal matrix $P$ and diagonal matrix $D$ such that $A = P^\intercal D P$.
Further:
- the diagonal entries of $D$ are the eigenvalues of $A$.
From Real Symmetric Positive Definite Matrix has Positive Eigenvalues:
- the diagonal entries of $D$ are positive.
We can therefore construct a real diagonal matrix $S$ by:
- $\paren S_{i j} = \begin{cases} \sqrt {\paren D_{i i} } & i = j \\ 0 & i \ne j \end{cases}$
From Product of Diagonal Matrices is Diagonal, we have:
- $\paren {S^2}_{i j} = \begin{cases} \paren D_{i i} & i = j \\ 0 & i \ne j \end{cases}$
so:
- $S^2 = D$
We also have:
| \(\ds \det S\) | \(=\) | \(\ds \prod_{i \mathop = 1}^n \sqrt {\paren D_{i i} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sqrt {\prod_{i \mathop = 1}^n \paren D_{i i} }\) | ||||||||||||
| \(\ds \) | \(>\) | \(\ds 0\) | as $\paren D_{i i} > 0$ for each $i$ |
We therefore have:
| \(\ds \map \det {P^\intercal S P}\) | \(=\) | \(\ds \map \det {P^\intercal} \det S \det P\) | Determinant of Matrix Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\det P}^2 \det S\) | Determinant of Transpose | |||||||||||
| \(\ds \) | \(=\) | \(\ds \det S\) | Determinant of Orthogonal Matrix is Plus or Minus One | |||||||||||
| \(\ds \) | \(>\) | \(\ds 0\) |
So from Matrix is Nonsingular iff Determinant has Multiplicative Inverse:
- $P^\intercal S P$ is nonsingular.
Let $C = P^\intercal S P$.
Then:
| \(\ds C^\intercal C\) | \(=\) | \(\ds \paren {P^\intercal S P}^\intercal P^\intercal S P\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds P^\intercal \paren {P^\intercal S}^\intercal P^\intercal S P\) | Transpose of Matrix Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds P^\intercal S^\intercal P P^\intercal S P\) | Transpose of Matrix Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds P^\intercal S^\intercal S P\) | as $P$ is orthogonal | |||||||||||
| \(\ds \) | \(=\) | \(\ds P^\intercal S^2 P\) | Diagonal Matrix is Symmetric | |||||||||||
| \(\ds \) | \(=\) | \(\ds P^\intercal D P\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds A\) |
As $C$ is nonsingular, the proof is complete.
$\Box$
Sufficient Condition
Let $A$ be a symmetric matrix such that:
- there exists an nonsingular matrix $C$ such that $A = C^\intercal C$.
Let $\mathbf v$ be a non-zero vector.
Then:
| \(\ds \mathbf v^\intercal A \mathbf v\) | \(=\) | \(\ds \mathbf v^\intercal C^\intercal C \mathbf v\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {C \mathbf v}^\intercal C \mathbf v\) | Transpose of Matrix Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {C \mathbf v} \cdot \paren {C \mathbf v}\) | Definition of Dot Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {C \mathbf v}^2\) | Dot Product of Vector with Itself | |||||||||||
| \(\ds \) | \(>\) | \(\ds 0\) | Euclidean Space is Normed Vector Space |
So $A$ is positive definite.
$\blacksquare$