Why is the matrix [1,0;0,0] not positive definite?

Question

If I take a vector v = [a,b], then isn't v.Mv = a^2, which is strictly greater than zero for all a and b not equal to zero?

"for every non zero vector": what if $a=0$ and $b=1$? – Math-fun Mar 07 '15 at 20:53 — Math-fun, Mar 07 '15 at 20:53

score 2 · Answer 1 · answered Mar 07 '15 at 20:47

2

Positive definite means $v^tMv > 0$ for all $v \neq 0$. Not so for your matrix $M$. E.g., $v^t = (0, 1)$.

answered Mar 07 '15 at 20:47

Simon S

26,524

score 1 · Answer 2 · answered Mar 07 '15 at 20:50

1

What you say is true. Yet the condition is for every non-zero vector. Note that $a$ can be $0$ while $v$ is not the zero-vector (when $b\neq 0$).

answered Mar 07 '15 at 20:50

quid

42,135

score 0 · Answer 3 · answered Mar 07 '15 at 20:52

For a matrix to be positive definite, you need $\mathbf{v}^TM\mathbf{v}>0$ for any $\mathbf{v}\neq\mathbf{0}$.

An arbitrary vector $\mathbf{v}=(a,b)$ will indeed give you $\mathbf{v}^TM\mathbf{v}=a^2$, but $a^2\ge 0$, i.e it is not necessary that $a^2\neq 0$. You could for instance pick the vector $\mathbf{v}^T=(0,1)$.

However, you matrix is positive semi-definite, since you have $\mathbf{v}^TM\mathbf{v}\ge 0$.

Why is the matrix [1,0;0,0] not positive definite?

3 Answers3