A very elegant and useful fact concerning self-adjoint transformations is the following minimax principle .
Theorem 1. Let \(A\) be a self-adjoint transformation on an \(n\) -dimensional inner product space \(\mathcal{V}\) , and let \(\lambda_{1}, \ldots, \lambda_{n}\) be the (not necessarily distinct) proper values of \(A\) , with the notation so chosen that \(\lambda_{1} \geq \lambda_{2} \geq \cdots \geq \lambda_{n}\) . If, for each subspace \(\mathcal{M}\) of \(\mathcal{V}\) , \[\mu(\mathcal{M})=\sup \big\{(A x, x): x \text { in } \mathcal{M},\|x\|=1\big\},\] and if, for \(k=1, \ldots, n\) , \[\mu_{k}=\inf \big\{\mu(\mathcal{M}): \operatorname{dim} \mathcal{M}=n-k+1\big\},\] then \(\mu_{k}=\lambda_{k}\) for \(k=1, \ldots, n\) .
Proof. Let \(\{x_{1}, \ldots, x_{n}\}\) be an orthonormal basis in \(\mathcal{V}\) for which \(A x_{i}=\lambda_{i} x_{i}\) , \(i=1, \ldots, n\) ( Section: Spectral theorem ); let \(\mathcal{M}_{k}\) be the subspace spanned by \(x_{1}, \ldots, x_{k}\) , for \(k=1, \ldots, n\) . Since the dimension of \(\mathcal{M}_{k}\) is \(k\) , the subspace \(\mathcal{M}_{k}\) cannot be disjoint from any \((n-k+1)\) -dimensional subspace \(\mathcal{M}\) in \(\mathcal{V}\) ; if \(\mathcal{M}\) is any such subspace, we may find a vector \(x\) belonging to both \(\mathcal{M}_{k}\) and \(\mathcal{M}\) and such that \(\|x\|=1\) . For this \(x=\sum_{i=1}^{k} \xi_{i} x_{i}\) we have \begin{align} (A x, x) &= \sum_{i=1}^{k} \lambda_{i}|\xi_{i}|^{2}\\ &\geq \lambda_{k} \sum_{i=1}^{k}|\xi_{i}|^{2} \\ &= \lambda_{k}\|x\|^{2}\\ &= \lambda_{k}, \end{align} so that \(\mu(\mathcal{M}) \geq \lambda_{k}\) .
If, on the other hand, we consider the particular \((n-k+1)\) -dimensional subspace \(\mathcal{M}_{0}\) spanned by \(x_{k}, x_{k+1}, \ldots, x_{n}\) , then, for each \(x=\sum_{i=k}^{n} \xi_{i} x_{i}\) in this subspace, we have (assuming \(\|x\|=1\) ) \begin{align} (A x, x) &= \sum_{i=k}^{n} \lambda_{i}|\xi_{i}|^{2}\\ &\leq \lambda_{k} \sum_{i=k}^{n}|\xi_{i}|^{2} \\ &= \lambda_{k}\|x\|^{2}\\ &= \lambda_{k}, \end{align} so that \(\mu(\mathcal{M}_{0}) \leq \lambda_{k}\) .
In other words, as \(\mathcal{M}\) runs over all \((n-k+1)\) -dimensional subspaces, \(\mu(\mathcal{M})\) is always \(\geq \lambda_{k}\) , and is at least once \(\leq \lambda_{k}\) ; this shows that \(\mu_{k}=\lambda_{k}\) , as was to be proved. ◻
In particular for \(k=1\) we see (using Section: Bounds of a self-adjoint transformation ) that if \(A\) is self-adjoint, then \(\|A\|\) is equal to the maximum of the absolute values of the proper values of \(A\) .
EXERCISES
Exercise 1. If \(\lambda\) is a proper value of a linear transformation \(A\) on a finite-dimensional inner product space, then \(|\lambda| \leq \|A\|\) .
Exercise 2. If \(A\) and \(B\) are linear transformations on a finite-dimensional unitary space, and if \(C=A B-B A\) , then \(\|1-C\| \geq 1\) . (Hint: consider the proper values of \(C\) .)
Exercise 3. If \(A\) and \(B\) are linear transformations on a finite-dimensional unitary space, if \(C=A B-B A\) , and if \(C\) commutes with \(A\) , then \(C\) is not invertible. (Hint: if \(C\) is invertible, then \(2\|B\| \cdot\|A\| \cdot\|A^{k-1}\| \geq k\|A^{k-1}\| /\|C^{-1}\|\) .)
Exercise 4.
- If \(A\) is a normal linear transformation on a finite-dimensional unitary space, then \(\|A\|\) is equal to the maximum of the absolute values of the proper values of \(A\) .
- Does the conclusion of (a) remain true if the hypothesis of normality is omitted?
Exercise 5. The spectral radius of a linear transformation \(A\) on a finite-dimensional unitary space, denoted by \(r(A)\) , is the maximum of the absolute values of the proper values of \(A\) .
- If \(f(\lambda)=((1-\lambda A)^{-1} x, y)\) , then \(f\) is an analytic function of \(\lambda\) in the region determined by \(|\lambda|<\frac{1}{r(A)}\) (for each fixed \(x\) and \(y\) ).
- There exists a constant \(K\) such that \(|\lambda|^{n}\|A^{n}\| \leq K\) whenever \(|\lambda|<\frac{1}{r(A)}\) and \(n=0,1,2, \ldots\) . (Hint: for each \(x\) and \(y\) there exists a constant \(K\) such that \(|\lambda^{n}(A^{n} x, y)| \leq K\) for all \(n\) .)
- \(\lim\sup_{n} \|A^{n}\|^{1 / n} \leq r(A)\) .
- \((r(A))^{n} \leq r(A^{n})\) , \(n=0,1,2, \ldots\) .
- \(r(A)=\lim_{n}\|A^{n}\|^{1 / n}\) .
Exercise 6. If \(A\) is a linear transformation on a finite-dimensional unitary space, then a necessary and sufficient condition that \(r(A)=\|A\|\) is that \(\|A^{n}\|=\|A\|^{n}\) for \(n=0,1,2, \ldots\) .
Exercise 7.
- If \(A\) is a positive linear transformation on a finite-dimensional inner product space, and if \(A B\) is self-adjoint, then \[|(A B x, x)| \leq\|B\| \cdot(A x, x)\] for every vector \(x\) .
- Does the conclusion of (a) remain true if \(\|B\|\) is replaced by \(r(B)\) ?