If an eigenvalue pair $\lambda_1 \cdot \lambda_2 < 0$ exists, then there is a nonzero vector $\vec{v}$ such that $A\vec{v}$ is orthogonal to $\vec{v}$ Suppose $A$ is an invertible, real symmetric $n\times n$ matrix. Prove if $A$ has at least one eigenvalue pair $\lambda_1, \lambda_2$ such that $\lambda_1\cdot \lambda_2 < 0$, then there exists a non-zero $\vec{v} \in \mathbb{R}^n$ such that $A\vec{v}$ is orthogonal to $\vec{v}$.
I know I need to show that there is a non-zero vector $\vec{v}$ such that $(\lambda_1\vec{v})\cdot \vec{v} = 0$ or $(\lambda_2\vec{v})\cdot\vec{v} = 0$, but I'm not really sure why this is true. If $A$ is symmetric, then $A^T = A$, but why does this make the question true?
 A: Hint: Assume $\lambda_1 >0, \lambda_2<0$ be eigenvalues of $A$ with corresponding eigenvectors $v_1,v_2$. Then, $A (c_1 v_1 + c_2 v_2) = c_1 \lambda_1 v_1 + c_2 \lambda_2 v_2$, and $<A (c_1 v_1 + c_2 v_2), c_1 v_1 + c_2 v_2> = c_1^2 \lambda_1 + c_2^2 \lambda_2$ since eigenvectors corresponding to different eigenvalues must be orthogonal (and thus linearly independent). Now, what should you pick c_1, c_2 to make the inner product zero? [ WLOG, you can take either $c_1$ or $c_2$ to be 1. ]
A: Let $v_1, v_2, \dots, v_n$ be an orthonormal basis of eigenvectors of $A$ (possible because $A$ is symmetric). Let $\lambda_1, \lambda_2$ etc. be the corresponding eigenvalues, and let the product of the first two be negative. Assume, WLOG that $\lambda_1 < 0 < \lambda_2$. 
Write $v = c_1 v_1 + c_2 v_2$, and we'll find $c_1,c_2$ so that $\langle Av, v \rangle = 0$.
Check that
$$
\langle Av, v \rangle = \langle c_1\lambda_1 v_1 + c_2\lambda_2v_2, c_1 v_1 + c_2 v_2 \rangle \\
= c_1^2 \lambda_1 |v_1|^2 + c_2^2 \lambda_2 |v_2|^2 = c_1^2 \lambda_1 + c_2^2 \lambda_2
$$
To make things easy, choose $c_1 = 1$, then find $c_2$ so that the above expression is zero (do you see why the assumption $\lambda_1 < 0 < \lambda_2$ makes this possible?)
A: Since $\mathbf{A}$ is symmetric,
it has an eigenvalue decomposition $\mathbf{A} = \mathbf{U} \mathbf{\Lambda} \mathbf{U}^{T}$, where $\mathbf{U}$ has orthonormal columns $\mathbf{u}_{1}, \dots, \mathbf{u}_{n}$, and $\Lambda$ is the diagonal matrix of eigenvalues.
Then
$$
\mathbf{A} = 
\mathbf{U} \mathbf{\Lambda} \mathbf{U}^{T}
=
\sum_{i=1}^{n} \lambda_{i}\mathbf{u}_{i}\mathbf{u}_{i}^{T}.
$$
Now, take any vector $\mathbf{v}$.
Then,
$$
\mathbf{w} = \mathbf{A}\mathbf{v} 
=
\sum_{i=1}^{n} \lambda_{i}\mathbf{u}_{i}\mathbf{u}_{i}^{T}\mathbf{v}
=
\sum_{i=1}^{n} \lambda_{i}(\mathbf{u}_{i}^{T}\mathbf{v})\mathbf{u}_{i}.
$$
You want to show that there exists a $\mathbf{v}$ such that $\mathbf{w} = \mathbf{A}\mathbf{v}$ is orthogonal to $\mathbf{v}$, or equivalently $\mathbf{w}^{T}\mathbf{v} = 0$.
But,
$$
\mathbf{w}^{T}\mathbf{v}
= \mathbf{v}^{T}\mathbf{A}^{T}\mathbf{v} 
= \mathbf{v}^{T}\mathbf{A}\mathbf{v} 
=
\sum_{i=1}^{n} \lambda_{i}(\mathbf{u}_{i}^{T}\mathbf{v})\mathbf{u}_{i}^{T}\mathbf{v}
=
\sum_{i=1}^{n} \lambda_{i}(\mathbf{u}_{i}^{T}\mathbf{v})^{2}.
$$
Note that $(\mathbf{u}_{i}^{T}\mathbf{v})^{2} >0$, for any $\mathbf{v}$ and $\mathbf{u}_{i}$. 
Exploiting the fact that $\lambda_{1}$ and $\lambda_{2}$ have opposite signs, you can carefully pick a $\mathbf{v}$ that makes the last sum equal to $0$, satisfying the desired property.
