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) ?
