# Tag Info

5

I have a guess as to the minimizer. Perhaps there's a neat proof that this happens to be the right answer. Note that $A$ has a singular value decomposition $$A = U\Sigma V^T$$ where $\Sigma$ is in this case a square, diagonal matrix whose diagonal elements are the (strictly positive) singular values of $A$, and $U,V$ are orthogonal matrices. My ...

2

Let us prove Omnomnomnom's suspicion. Fix $A \in M_n(\mathbb{R})$. Following the OP's observation, we want to maximize the linear functional $\varphi \colon M_n(\mathbb{R}) \rightarrow \mathbb{R}$ given by $\varphi(X) = \mathrm{tr}(A^tX)$ subject to the constraint $X^t X = I$. First, let us assume that $A$ is a diagonal matrix with non-negative entries on ...

2

For simply diagonally dominant it is not true. The counterexample is $$A=\left[\matrix{-1 & 1\\1 & -1}\right].$$ For strictly diagonally dominant: assume that the matrix $A=\{a_{ij}\}$ satisfies $a_{ij}\ge 0$, $i\ne j$ $a_{ii}+\sum_{j\ne i}a_{ij}<0$ then by Gerschgorin's theorem all the eigenvalues of $A$ have negative real parts, i.e. the ...

2

Show that T is a linear transformation A linear transformation $f: V \to W$ has two properties: Homogeneity of degree $1$: $f(\alpha x) = \alpha f(x)$ Additivity: $f(x+y) = f(x)+f(y)$ These can be tested at the same time by confirming that $f(\alpha x + y) = \alpha f(x) + f(y)$ for all $x,y \in V$ and $\alpha$ in your base field (probably $\Bbb R$ ...

1

Let's have a look at this inductively: start with the easy $$d_2=\begin{vmatrix} 1 & 1 \\ a & b \end{vmatrix} = (b-a).$$ Subtracting the first column from the second, we can write this as $$d_2=\begin{vmatrix} 1 & 1 \\ a & b \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ a & b-a \end{vmatrix}.$$ Now suppose we have $n$ columns:  ...

1

Since $D$ and $E$ are conjugate by $A$ (i.e., $E = ADA^{-1}$), they must have the same eigenvalues with eigenspaces of the same same dimension: if $Dx = \lambda x$, then $E(Ax) = ADx = \lambda Ax$. It follows from this that the diagonal entries of $E$ and $D$ are the same, in a possibly different order. If $D$ has diagonal entries $d_i$, consider ...

1

$H$ is simply the Cartesian product of groups $G_1 \times G_2$, and the irreducible representations turn out to be tensor products of the form $W_1 \otimes W_2$ where $W_i$ is an irreducible representation of $G_i$. The particular case of $3 \times 3$ matrices turns out to be the representation $V \otimes V^*$ where $V$ is the standard $3$-dimensional real ...

1

I know that online sources such as Wikipedia and Wolfram just state without any proof or extended discussions that the matrix exponential is well-defined and converges for any square matrix. Every matrix has an element of maximal size. $($Obviously, if anything can cause divergence, it's that one$).~$ Let its absolute value be $M.~$ So let us construct ...

1

One of the approaches: Let $S$ be a space of number sequences. Consider your matrices $M_{n_1,n_2}$ as linear operators on $S$ as follows: take a sequence $b$, cut off first $n_2$ entries, act on them by your matrix $M_{n_1,n_2}$, obtain $n_1$ entries, then project them on $S$ by filling the tail of the number sequence with zeros. If you introduce a norm ...

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