With these constraints, your matrix $\lambda$ (I suggest you use the notation $\Lambda$ instead, because $\lambda$ usually denotes a scalar) must be of the form $c \mathbf{1}\mathbf{1}^T$, where $\... View answer 2 votes Yes, that is true in general. First, note that by definition the left nullspace of$A$is the orthogonal complement of its column space (which, by the way, is unique, and so we say "the column ... View answer Accepted answer 1 votes Your guess is right, but the explanation is not entirely accurate. First, note that saying$A x = 0$has only the trivial solution is actually equivalent to saying that the nullspace of$A$only ... View answer Accepted answer 1 votes It all boils down to the elementary divisors of your matrix, because two matrices are similar if and only if they have the same elementary divisors. In other words, similarity transformations always ... View answer 1 votes Well, first note that every continuous linear form in$\mathbb{R}^n \rightarrow \mathbb{R}$is represented by a unique vector in$\mathbb{R}^n$(this is the Riesz representation theorem), e.g.,$\...
To go from (2) to (3), first note that the rightmost summation is a linear combination of vectors $\tilde{\mathbf{x}}_n$, with coefficients with happen to be scalar products of the form \$\beta^T \...