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Let $A$ be a real symmetric $n\times n$ matrix, and let $X$ be the set of the eigenvalues of $A$. I was wondering if $$X = \{ \langle Ax , x\rangle : x\in\mathbb{R}^n,\|x\|=1 \}.$$ I think that the equality holds:

Let $\lambda\in X$. Since $A$ is symmetric, we have that $\lambda$ is real. Then, there exists a unit vector $v$ in $\mathbb{R}^n$ such that $Av = \lambda v$. Thus, $$\langle Av,v\rangle = \langle\lambda v,v\rangle = \lambda\langle v,v\rangle = \lambda \|v\|^2 = \lambda.$$ It follows that $\lambda \in \{ \langle Ax , x\rangle : x\in\mathbb{R}^n,\|x\|=1 \}$.

Now, let $\alpha\in \{ \langle Ax , x\rangle : x\in\mathbb{R}^n,\|x\|=1 \}$. Then, there exists a unit vector $u$ in $\mathbb{R}^n$ such that $\alpha = \langle Au,u\rangle$, which implies that $u^TAu = \alpha$. Hence, $\alpha u = uu^TAu = \|u\|^2Au = Au$. Therefore, $\alpha$ is an eigenvalue of $A$.

Is my reasoning correct? Thanks in advance.

Lord Vader
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  • No, .why would $x$ always be an eigenvector – Andrew Aug 26 '23 at 01:11
  • Please work through a $2\times 2$ diagonal matrix example where there are two distinct eigenvalues -- i suggest $0$ and $1$. – user8675309 Aug 26 '23 at 01:29
  • @Andrew Yes, you're right. The question arised beacause I'm trying to understand the following result: If $A$ is symmetric and $\lambda$ is the greatest eigenvalue of $A$, then $\lambda = \max{\langle Ax,x\rangle:x\in\mathbb{R}^n,|x|=1}$, and the maximum is attained only on the eigenvectors of $\lambda$. I know how to prove that the values of the given set are bounded by $\lambda$, but how can I show that the only unit vectors $x$ such that $\langle Ax,x\rangle=\lambda$ are eigenvalues of $\lambda$? – Lord Vader Aug 26 '23 at 01:32
  • Maximum and minimum, yes, but (by intermediate value theorem) not the rest. What you’re trying to show follows immediately from the spectral theorem. – Ted Shifrin Aug 26 '23 at 01:40
  • Your last paragraph is Incorrect. Look carefully at $uu^\top$. – Ted Shifrin Aug 26 '23 at 01:49
  • @TedShifrin thanks for your answer. I have written another post to improve my question. May you read it? https://math.stackexchange.com/questions/4758768/clarification-about-the-answer-of-a-post – Lord Vader Aug 26 '23 at 01:51
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    You should not post essentially the same question multiple times. I already answered your question: Use the spectral theorem. – Ted Shifrin Aug 26 '23 at 01:56

2 Answers2

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The set $X$ has at most $n$ elements. Meanwhile, the set $$ W(A)= \{ \langle Ax , x\rangle : x\in\mathbb{R}^n,\|x\|=1 \}, $$ which is called the numerical range of $A$ is uncountably infinite as long as $A$ is not a scalar multiple of the identity.

The numerical range is always convex (this is the well-known Hausdorff–Toeplitz theorem), and it always contains the spectrum.

In summary, your equality can only occur if $A$ is a scalar multiple of the identity.

Martin Argerami
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    In the real case there's no need to appeal to the Hausdorff-Toeplitz theorem, is there? Here the numerical range is the image of $S^{n-1}$ under a continuous map into $\mathbb{R}$, which is compact and connected, so it must be a closed interval. – Qiaochu Yuan Aug 30 '23 at 22:43
  • Indeed. $ \ \ \ \ $ – Martin Argerami Aug 30 '23 at 23:20
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Now, let $\alpha\in \{ \langle Ax , x\rangle : x\in\mathbb{R}^n,\|x\|=1 \}$. Then, there exists a unit vector $u$ in $\mathbb{R}^n$ such that $\alpha = \langle Au,u\rangle$, which implies that $u^TAu = \alpha$. Hence, $\alpha u = uu^TAu = \|u\|^2Au = Au$. Therefore, $\alpha$ is an eigenvalue of $A$.

This is incorrect. It's $u^T u$ that is equal to $\| u \|^2$; the multiplication $uu^T$ going the other way is an $n \times n$ matrix, not a scalar. As Andrew says in the comments, there's no reason for $u$ to be an eigenvector; by hypothesis $u$ can be an arbitrary unit vector.

The numerical range $\{ \langle Ax, x \rangle : x \in \mathbb{R}^n, \| x \| = 1 \}$ is in fact always the closed interval $[\lambda_n, \lambda_1]$, where $\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_n$ are the eigenvalues of $A$ in decreasing order. This follows pretty straightforwardly from the spectral theorem.

I know how to prove that the values of the given set are bounded by $\lambda$, but how can I show that the only unit vectors $x$ such that $\langle Ax, x \rangle = \lambda$ are eigenvalues of $\lambda$?

Write $f(x) = \langle Ax, x \rangle$. If $v$ is such that $f(v) = \lambda_1$ attains its maximum value then the derivative $df_v$ must be equal to zero. This derivative is given by the linear map

$$w \mapsto \langle Av, w \rangle + \langle Aw, v \rangle = 2 \langle Av, w \rangle$$

where the condition that $w$ lies in the tangent space to $v$ on the unit sphere gives that $\langle v, w \rangle = 0$. So we get that if $w$ is orthogonal to $v$ then it must also be orthogonal to $Av$; this gives that $Av$ is a scalar multiple of $v$ as desired (and hence that the maximum value $\lambda_1$ really is an eigenvalue). This argument does not rely on the spectral theorem and in fact is the beginning of a proof of the spectral theorem.

Qiaochu Yuan
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