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edited Jan 14 at 0:03

Let me give it a try. This one only uses the existence of a maximum in a compact set, and the Cauchy-Schwarz inequality.

Let $T$ be a selfadjoint operator in a finite dimensional inner product space.

Claim: $T$ has an eigenvalue $\pm\|T\|$.

Proof: Let $v$ in the unit sphere be such that $\|Tv\|$ attains its maximum value $M=\|T\|$. Let $w$ also in the unit sphere be such that $Mw=Tv$ (which is like saying that $w=\frac{Tv}M$, except in the trivial case $T=0$).

This implies that $\langle w,Tv\rangle=M$. In fact, the only way that to unit vectors $v$ and $w$ can satisfy this equation is to have $Mw=Tv$. (Since we know that $\|w\|=1$ and $\|Tv\|\leq M$, the Cauchy-Schwarz inequality tells us that $|\langle w,Tv\rangle\|\leq M$, and the equality case is only attainable when $Tv$ is a scalar multiple of $w$, being $M$ the only possible value of the scalar.)

But by selfadjointness of $T$, we also know that $\langle v,Tw\rangle=M$, so that $Mv=Tw$.

Now, one of the two vectors $v\pm w$ is nonzero, and we can compute

$T(v\pm w)=Tv\pm Tw=Mw\pm Mv=M(w\pm v)=\pm M(v\pm w)$.

This concludes the proof that $\pm\|T\|$ is eigenvalue with eigenvector $v\pm w$. The reality of the other eigenvalues can be proved by induction, restricting to $(v\pm w)^\bot$ as in the usual proof of the spectral theorem.

Remark: The proof above works with real or complex spaces, and also for compact operators in Hilbert spaces.

Comment: I would like to know if this proof can be found in the literature. I obtained it while trying to simplify a proof of the fact that if $T$ is a bounded selfadjoint operator, then $\|T\|=\sup_{\|v\|\leq 1} \langle Tv,v\rangle$ (as found, for example, on p.32 of Conway J.B., "An Introduction to Functional Analysis"). In the case of non-compact operators, one can only prove that $T$ has as an approximate eigenvalue one of the numbers $\pm\|T\|$. The argument is similar to the one above, but knowledge of the equality case of Cauchy-Schwarz is not enough. One has to know that near-equality implies near-dependence. More precisely, let $v$ be a fixed unit vector, $M\geq 0$ and $\varepsilon\in[0,1]$. \varepsilon\in[0,M]$. If $z$ is a vector with $\|z\|\leq M$ such that $|\langle v,z\rangle|\geq \sqrt{M^2-\epsilon^2}$, then it can be proved that $z$ is within distance $\varepsilon$ of $\langle v,z\rangle v$.

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edited Jan 13 at 23:57

Marcos Cossarini
713●6●8

Let me give it a try. This one only uses the existence of a maximum in a compact set, and the Cauchy-Schwarz inequality.

Let $T$ be a selfadjoint operator in a finite dimensional inner product space.

Claim: $T$ has an eigenvalue $\pm\|T\|$.

But by selfadjointness of $T$, we also know that $\langle v,Tw\rangle=M$, so that $Mv=Tw$.

Now, one of the two vectors $v\pm w$ is nonzero, and we can compute

$T(v\pm w)=Tv\pm Tw=Mw\pm Mv=M(w\pm v)=\pm M(v\pm w)$.

Remark: The proof above works with real or complex spaces, and also for compact operators in Hilbert spaces.

Comment: I would like to know if this proof can be found in the literature. I obtained it while trying to simplify a proof of the fact that if $T$ is a bounded selfadjoint operator, then $\|T\|=\sup_{\|v\|\leq 1} \langle Tv,v\rangle$ (as found, for example, on p.32 of Conway J.B., "An Introduction to Functional Analysis"). In the case of non-compact operators, one can only prove that $T$ has as an approximate eigenvalue one of the numbers $\pm\|T\|$. The argument is similar to the one above, but knowledge of the equality case of Cauchy-Schwarz is not enough. One has to know that near-equality implies near-dependence. More precisely, let $v$ be a fixed unit vector, $M\geq 0$ and $\varepsilon\in[0,1]$. If $z$ is a vector with $\|z\|\leq M$ such that $|\langle v,z\rangle|\geq M\sqrt{1-\epsilon^2}$, \sqrt{M^2-\epsilon^2}$, then it can be proved that $z$ is within distance $\varepsilon M$ \varepsilon$ of $\langle v,z\rangle v$.

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edited Jan 13 at 6:01

Marcos Cossarini
713●6●8

Let me give it a try. This one only uses the existence of a maximum in a compact set, and the Cauchy-Schwarz inequality.

Let $T$ be a selfadjoint operator in a finite dimensional inner product space.

Claim: $T$ has an eigenvalue $\pm\|T\|$.

But by selfadjointness of $T$, we also know that $\langle v,Tw\rangle=M$, so that $Mv=Tw$.

Now, one of the two vectors $v\pm w$ is nonzero, and we can compute

$T(v\pm w)=Tv\pm Tw=Mw\pm Mv=M(w\pm v)=\pm M(v\pm w)$.

Remark: The proof above works with real or complex spaces, and also for compact operators in Hilbert spaces.

Comment: I would like to know if this proof can be found in the literature. I obtained it while trying to simplify a proof of the fact that if $T$ is a bounded selfadjoint operator, then $\|T\|=\sup_{\|v\|\leq 1} \langle Tv,v\rangle$ (as found, for example, on p.32 of Conway J.B., "An Introduction to Functional Analysis"). In the case of non-compact operators, one can only prove that $T$ has as an approximate eigenvalue one of the numbers $\pm\|T\|$. The argument is similar to the one above, but knowledge of the equality case of Cauchy-Schwarz is not enough. One has to know that near-equality implies near-dependence. More precisely, let $v$ be a fixed unit vector, $M\geq 0$ and $\varepsilon\in[0,1]$. If $z$ is a vector with $\|z\|\leq M$ such that $|\langle v,z\rangle|\geq M\sqrt{1-\epsilon^2}$, then it can be proved that $z$ is within distance $\varepsilon M$ of $v\langle v,z\rangle$\langle v,z\rangle v$.

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edited Jan 12 at 21:36

Marcos Cossarini
713●6●8

Let me give it a try. This one is nice and only uses the existence of a maximum in a compact set, and the Cauchy-Schwarz inequality.

Let $T$ be a selfadjoint operator in a finite dimensional inner product space.

Claim: $T$ has an eigenvalue $\pm\|T\|$.

But by selfadjointness of $T$, we also know that $\langle v,Tw\rangle=M$, so that $Mv=Tw$.

Now, one of the two vectors $v\pm w$ is nonzero, and we can compute

$T(v\pm w)=Tv\pm Tw=Mw\pm Mv=M(w\pm v)=\pm M(v\pm w)$.

Remark: The proof above works with real or complex spaces, and also for compact operators in Hilbert spaces.

Comment: I would like to know if this proof can be found in the literature. I obtained it while trying to simplify a proof of the fact that if $T$ is a bounded selfadjoint operator, then $\|T\|=\sup_{\|v\|\leq 1} \langle Tv,v\rangle$ (as found, for example, on p.32 of Conway J.B., "An Introduction to Functional Analysis"). In the case of non-compact operators, one can only prove that $T$ has as an approximate eigenvalue one of the numbers $\pm\|T\|$. The argument is similar to the one above, but knowledge of the equality case of Cauchy-Schwarz is not enough. One has to know that near-equality implies near-dependence. More precisely, let $v$ be a fixed unit vector, $M\geq 0$ and $\varepsilon\in[0,1]$. If $z$ is a vector with $\|z\|\leq M$ such that $|\langle v,z\rangle|\geq M\sqrt{1-\epsilon^2}$, then it can be proved that $z$ is within distance $\varepsilon M$ of $v\langle v,z\rangle$.

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answered Jan 12 at 20:42

Marcos Cossarini
713●6●8