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Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite Eigenvectors with distinct eigenvalues are linearly independant, but what about the case where eigenvalues are repeated. An example is where we have matrix A = $$ beginmatrix 1 & 0 \ 0 & 1 \ endmatrix $$ with eigenvalues $λ$ = 1 which is repeated twice so spectrum A is 1(2). And we find that the associated augments matrix is $$ beginmatrix 0 & 0 \ 0 & 0 \ endmatrix $$ And because neither component of the vector is contrained, the eigenvectors is of the form [u v]^T. We observe that [u v]^T = u[1 0]^T + v[0 1]^T. Where [1 0]^T + [0 1]^T is linearly independant. So from here how does this actually show that eigenvectors with repeated eigenvalues are not linearly dependant or otherwise? linear-algebra matrices eigenvalues-eigenvectors share | cite | improve this question asked Aug 25 at 11:20 Randy Rogers 72 6 From [u v]^T = u[1 0]^T + v...

Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite For the plane $z=50$ on the surface $z=75-x^2-y^2$, the contour lines is the circle $x^2+y^2=25$ in the xy-plane. My question is why we say "contour lines" not "contour curve", also we use "lines" not "line" ? multivariable-calculus share | cite | improve this question edited Aug 25 at 11:43 user144410 744 1 5 19 asked Aug 25 at 11:28 Adesh Sharma 2 2 These terms are used interchangeably – MPW Aug 25 at 11:30 if the graph of $z$ is chopped with the plane $z=50$, the resulting circle $x^2+y^2=25$ is the contour curve. It is a single curve and not a (straight) line. If the objective function $z$ is a linear function (of degree one), then there will be contour line (if cut by one plane) or contour lines (if cut by several planes). – farruhota Aug 25 at 12:13 add a comment  |Â...

Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite According to official QEMU documentation: When used as a virtualizer, QEMU achieves near native performance by executing the guest code directly on the host CPU. My question is how is this behavior different from the programs we call hypervisors (e.g. VirtualBox, KVM etc); don't they also run the "guest code" on the "host CPU"? where else would they run it on? Update : If the image below (source) is correct, then it turns out that QEMU runs applications (and not entire OSs like hypervisors) but performs translations between different architecture types (e.g. a program written for ARM can run on x86) Given the above, is the case that for one to run: a) an entire operating system for a specific architecture (e.g ARM) b) on a host with a different architecture (e.g amd64) a co existence is needed of both QEMU (to perform the cross-architecture mapping) and KVM (to act as hardware hypervisor) ...