Solutions of Selected Theoretical Exercises, Linear Algebra
... as A = S + K, where S is symmetric and K is skew-symmetric. Solution. Suppose such a decomposition exists. Then AT = S T + K T = S − K so that A + AT = 2S and A − AT = 2K. Now take S = 21 (A + AT ) and K = 21 (A − AT ). One verifies A = S + K, S = S T and K T = −K similarly to the previous Exercise ...
... as A = S + K, where S is symmetric and K is skew-symmetric. Solution. Suppose such a decomposition exists. Then AT = S T + K T = S − K so that A + AT = 2S and A − AT = 2K. Now take S = 21 (A + AT ) and K = 21 (A − AT ). One verifies A = S + K, S = S T and K T = −K similarly to the previous Exercise ...
Ordinary Differential Equations: A Linear Algebra
... scalar problems. We almost immediately go the case of the homogeneous problem being constant coefficient, and derive the homogeneous solution via an expansion in terms of eigenvectors. From a pedagogical perspective I find (and my students seem to agree) this to be a natural way to see how the eigen ...
... scalar problems. We almost immediately go the case of the homogeneous problem being constant coefficient, and derive the homogeneous solution via an expansion in terms of eigenvectors. From a pedagogical perspective I find (and my students seem to agree) this to be a natural way to see how the eigen ...
