Real spectrum nonempty iff exists $(n-1)$-dimensional invariant subspace

Question

I'm trying to prove the following:

Let $V$ be a real finite-dimensional vector space, $\dim V = n$, and $A : V \to V$ a linear map.

$A$ has a nonempty (real) spectrum if and only if there exists an $A$-invariant subspace $W\leq V$ such that $\dim W = n-1$.

$\fbox{$\impliedby$}$

Assume there exists an $(n-1)$-dimensional and $A$-invariant subspace $W \leq V$. Pick a basis $\{w_1, \ldots, w_{n-1}\}$ for $W$, and extend it to a basis $\{w_1, \ldots, w_{n-1}, z\}$ for $V$.

The matrix representation of $A$ in this basis is:

$$\left[\begin{array}{cccc|c} a_{1,1} & a_{1,2} & \ldots & a_{1,n-1} & a_{1,n}\\ a_{2,1} & a_{2,2} & \ldots & a_{2,n-1} & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_{n-1,1} & a_{n-1,2} & \ldots & a_{n-1,n-1} & a_{n-1,n}\\ \hline 0 & 0 & \cdots & 0 & a_{n,n} \end{array}\right]$$

We see that $Az = a_{n,n}z$. Furthermore, $a_{n,n} \in \mathbb{R}$ and $z \ne 0$ imply $a_{n,n} \in \sigma(A)$.

$\fbox{$\implies$}$

Assume $\sigma(A)\ne\emptyset$.

Take $\lambda \in \sigma(A)$. Then there exists $x \in V$, $x\ne 0$ such that $Ax = \lambda x$. Extend the linearly independent set $\{x\}$ to a basis $\{x, x_2, \ldots, x_n\}$ of $V$.

The matrix representation of $A$ in this basis is:

$$\left[\begin{array}{c|cccc} \lambda & a_{1,2} & \ldots & a_{1,n-1} & a_{1,n} \\ \hline 0 & a_{2,2} & \ldots & a_{2,n-1} & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & a_{n-1,2} & \ldots & a_{n-1,n-1} & a_{n-1,n}\\ 0 & a_{n,2} & \ldots & a_{n,n-1} & a_{n,n} \end{array}\right]$$

Is there a way to transform $[\{x_2, \ldots, x_n\}]$ to an invariant subspace (so that the first row has zeroes after $\lambda$)?

Another attempt:

Consider the standard inner product $\langle\cdot,\cdot\rangle$ on $\mathbb{R}^n$. Since $$\sigma(A^T) = \sigma(A^*) = \overline{\sigma(A)} = \sigma(A)$$ we have $\lambda \in \sigma(A^T)$. Pick $w \ne 0$ such that $A^T w = \lambda w$.

We claim that $\{w\}^\perp$ is $A$-invariant:

Take $y \in \{w\}^\perp$. We have:

$$\langle Ay,w\rangle = \langle y,A^T w\rangle = \langle y,\lambda w\rangle = \lambda \langle y,w\rangle = 0$$

This implies $Aw \in \{w\}^\perp$.

Since $\dim \{w\}^\perp = n-1$, we have found the desired subspace.

Is my proof correct? Is there a way to finish $\fbox{$\implies$}$ without introducing the inner product?

Answer 1

HINT:

1. The spectrum of $A$ and $A^{t}$ are the same.

2. $A$ has an invariant $n-1$ dim subspace if and only if $A^{t}$ has an invariant $1$ dim subspace.

You don't need the inner product, just transpose some equality $A = U B U^{-1}$.