Question about inner product space and positive definite matrix

Question

I was working on problems in a textbook for my intro linear algebra class, and one of the questions asked about showing that if we have some real positive definite matrix, A, then $\lt u,v\gt=u^{T}$$Av$ is an inner product on $\mathbb R^{n}$.

So I worked through it and it seemed not bad, I just showed that for all vectors it satisfied the linear property, symmetric property and positive definite property. And my solution seemed to be similar to the books approach. However it had a follow up problem that I am a little confused about. The question asked, using that same function with $$A= \begin{bmatrix} 1 & 3 \\ 3 & 9 \\ \end{bmatrix} $$to determine if it still an inner product.

So I am a little confused. I computed the null-space to be $N(A)=span{(-3,1)^{T}}$. But I am not sure if this is the right approach. Would this be telling me that the matrix may not be positive definite? I know because also A is symmetric so we have $ A=A^{T}$ and $N(A)$=$row(A)$$\bot$=$col(A)$.

Anyways, any suggestions or help/answers would be appreciated. Thanks all!

Answer 1

It can't be definite positive because all its eigenvalues should all be positive, and it is a singular matrix, hence $0$ is an eigenvalue for $A$ (the other one is $10$).

Without eigenvalues.

The matrix is singular and $u=\begin{bmatrix}3\\-1\end{bmatrix}\in \ker A$, hence $$^{\mathrm t}u A u=\begin{bmatrix}3&-1\end{bmatrix}\begin{bmatrix}0\\0\end{bmatrix}=0.$$

Answer 2

Definiteness of an inner product means that if $\langle u, u\rangle=0$, then $u=0$. For the choice of matrix $A$ as above, if you choose the vector $u$ in the null space of $A$, you can see that the definiteness is violated.

Answer 3

The matrix $A$ is definite positive if

$$x^T Ax>0,\; \forall x\ne0$$ so if $\ker A\ne\{0\}$ then clearly $A$ doesn't satisfy the above inequality by taking $0\ne x\in\ker A$.

Answer 4

If you note that the eigenvalues of $A$ are $0$ and $10$, then it cannot be positive definite.

Otherwise, you can take an arbitrary vector $u=(x,y)^T$, and compute $\langle u,u\rangle=u^TAu$:

$$u^TAu=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&3\\3&9\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=(x+3y)^2\ge 0$$

Therefore the matrix is positive semi-definite.