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 an invertible 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 Invertible iff Determinant has Multiplicative Inverse:

$P^\intercal S P$ is invertible.

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 invertible, the proof is complete.

$\Box$


Sufficient Condition

Let $A$ be a symmetric matrix such that:

there exists an invertible 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.

$\blacksquare$

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