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Given two symmetric matrices A and B with same dimensions, if $\lambda_{max}(A)$ is biggest eigenvalue of A, $\lambda_{min}(B)$ is the smallest eigenvalue of B, does the following holds true:

$$A \leq B \Rightarrow \lambda_{max}(A)\leq \lambda_{min}(B) \ \ ?$$

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    What do you mean by $A \leq B$ ? That $\forall i,j, \ a_{i,j} \leq b_{i,j}$ ? – 2017-01-28
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    @jeanMarie, that mean $*A*-*B* \leq 0$ – 2017-01-28
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    Do you want to have the ineqaulity elemtwise (every element is larger) or do you want to have $A-B$ is positive semidefinite? – 2017-01-28
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    I want to have A-B is negative semidefinite – 2017-01-28

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$\lambda_{max}(A) \leq \lambda_{min}(B)$ is a sufficient property for the so-called "Loewner order".

See (https://en.wikipedia.org/wiki/Loewner_order)

But is is not a necessary condition. Here is a counterexample; take:

$$A=\pmatrix{1&1\\1&2} \ \ \text{and} \ \ B=\pmatrix{2&2\\2&3}.$$

We have: $A \leq B$ for the Loewner order because $B-A=\pmatrix{1&1\\1&1}$ is semi-positive-definite (with eigenvalues $0$ and $2$).

The spectra of $A$ and $B$ are resp.

$$\{0.3819,2.6180\} \ \ \text{and} \ \ \{0.4384, 4.5616\}.$$

but without having $\lambda_{max}(A) \leq \lambda_{min}(B).$

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    but the postulat of the links means that λmax(A)≤λmin(B)⇒A≤B, and me I ask for the reciproque – 2017-01-28