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Usually a quantity of a matrix is defined as the eigenvalues of the matrix. If so, how can anyone express continuous values, as in Schrodinger picture, into a matrix?

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A matrix normally refers to a collection of matrix elements $M_{ij}$ where the indices $i,j$ take values in a discrete, or even finite, set. But one may generalize it so that the indices become effectively continuous, like $M(x,x')$. This corresponds to a "continuous basis" of the Hilbert space. In fact, the same infinite-dimensional Hilbert space may have both continuous and discrete bases, so the descriptions are equivalent. A more general word than a matrix is an operator. The operator $x:\psi(x)\mapsto x\psi(x)$ has continuous eigenvalues. It's still a matrix in any discrete basis. – Luboš Motl Sep 27 '12 at 14:26
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@Luboš that could be an answer – David Zaslavsky♦ Sep 27 '12 at 17:11
One should also be careful about the expectation value of an observable vs the eigenvalues. Such failures befall even the modern day greats and lead them (usually temporarily) to bizarre conclusions such as violation of Lorentz invariance at small scales. – genneth Oct 12 '12 at 20:38
This is what Dirac spends several chapters of his book explaining. It's the Dirac delta function, and distribution theory. – Ron Maimon Oct 13 '12 at 5:06

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A finite-dimensional matrix always has a discrete spectrum. For continuous quantities one needs operators with a continuous spectrum, which therefore must act on an infinite-dimensional space.

Examples are the multiplication operator or the differentiation operator in the space of sufficiently nice square integrable functions of a real variable. (This restriction is needed as they are only densely defined in the Hilbert space.)

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Continuous observables are represented by infinite matrices. E.g. the position $\mathbf{x}$ and momentum $\mathbf{p}$ matrices associated to the continuous basis $|x\rangle$ and $|p\rangle$ are infinite.

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