Positive definite matrix

Examples

• The identity matrix${\displaystyle I={\begin{bmatrix}1&0\\0&1\end{bmatrix}}}$ is positive-definite (and as such also positive semi-definite). It is a real symmetric matrix, and, for any non-zero column vector z with real entries a and b, one has

  $${\displaystyle \mathbf {z} ^{\mathsf {T}}I\mathbf {z} ={\begin{bmatrix}a&b\end{bmatrix}}{\begin{bmatrix}1&0\\0&1\end{bmatrix}}{\begin{bmatrix}a\\b\end{bmatrix}}=a^{2}+b^{2}.}$$ Seen as a complex matrix, for any non-zero column vector z with complex entries a and b one has $${\displaystyle \mathbf {z} ^{*}I\mathbf {z} ={\begin{bmatrix}{\overline {a}}&{\overline {b}}\end{bmatrix}}{\begin{bmatrix}1&0\\0&1\end{bmatrix}}{\begin{bmatrix}a\\b\end{bmatrix}}={\overline {a}}a+{\overline {b}}b=|a|^{2}+|b|^{2}.}$$

  Either way, the result is positive since ${\displaystyle \mathbf {z} }$ is not the zero vector (that is, at least one of ${\displaystyle a}$ and ${\displaystyle b}$ is not zero).
• The real symmetric matrix $${\displaystyle M={\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}}}$$ is positive-definite since for any non-zero column vector z with entries a, b and c, we have $${\displaystyle {\begin{aligned}\mathbf {z} ^{\mathsf {T}}M\mathbf {z} =\left(\mathbf {z} ^{\mathsf {T}}M\right)\mathbf {z} &={\begin{bmatrix}(2a-b)&(-a+2b-c)&(-b+2c)\end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}\\&=(2a-b)a+(-a+2b-c)b+(-b+2c)c\\&=2a^{2}-ba-ab+2b^{2}-cb-bc+2c^{2}\\&=2a^{2}-2ab+2b^{2}-2bc+2c^{2}\\&=a^{2}+a^{2}-2ab+b^{2}+b^{2}-2bc+c^{2}+c^{2}\\&=a^{2}+(a-b)^{2}+(b-c)^{2}+c^{2}\end{aligned}}}$$ This result is a sum of squares, and therefore non-negative; and is zero only if ${\displaystyle a=b=c=0,}$ that is, when ${\displaystyle \mathbf {z} }$ is the zero vector.
• For any real invertible matrix${\displaystyle A,}$ the product ${\displaystyle A^{\mathsf {T}}A}$ is a positive definite matrix (if the means of the columns of A are 0, then this is also called the covariance matrix). A simple proof is that for any non-zero vector ${\displaystyle \mathbf {z} ,}$ the condition ${\displaystyle \mathbf {z} ^{\mathsf {T}}A^{\mathsf {T}}A\mathbf {z} =(A\mathbf {z} )^{\mathsf {T}}(A\mathbf {z} )=\|A\mathbf {z} \|^{2}>0,}$ since the invertibility of matrix ${\displaystyle A}$ means that ${\displaystyle A\mathbf {z} \neq 0.}$
• The example ${\displaystyle M}$ above shows that a matrix in which some elements are negative may still be positive definite. Conversely, a matrix whose entries are all positive is not necessarily positive definite, as for example $${\displaystyle N={\begin{bmatrix}1&2\\2&1\end{bmatrix}},}$$ for which ${\displaystyle {\begin{bmatrix}-1&1\end{bmatrix}}N{\begin{bmatrix}-1&1\end{bmatrix}}^{\mathsf {T}}=-2<0.}$