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4 votes
Accepted

Iterative scheme to make matrice inverses of each other

Note that $$ X_{i+1} Y_{i+1} = (9 X_i Y_i - 6 (X_i Y_i)^2 + (X_i Y_i)^3)/4$$ so $$ X_{i+1} Y_{i+1} - I = \frac{(X_i Y_i - 4 I)(X_i Y_i - I)^2}{4} $$ and for any submultiplicative matrix norm $$ \|X_{i+...
Robert Israel's user avatar
4 votes
Accepted

If $[X,Y]=Y$ must $Y$ have $0$ as the only eigenvalue?

You can prove it using the following identity for the matrix exponential $$ e^ABe^{-A}= I+[A,B]+\frac12[A,[A,B]]+\frac16[A,[A,[A,B]]]+\ldots\,, $$ which shows that $$ e^{-tX} Ye^{tX} = e^{-t}Y\,. $$ ...
Ben's user avatar
  • 844
2 votes
Accepted

To prove or disprove $MM^t=\alpha I_{m\times m}$,where $M$ is an $m\times n$ matrix of rank $m$ with $m<n$.

Let call $M=(a_{ij})_{i \le m, j \le n}$, then the product on the LHS is given by $$ [x^tMM^tx]_{i}=\left(\sum_{k=1}^n a_{ik}^2\right) (x_i)^2+\sum_{j \ne i}^m \left(\sum_{k=1}^n a_{ik} a_{jk}\right) ...
Marco's user avatar
  • 2,615
2 votes
Accepted

Linear matrix equation with stacked matrices and repeated unknowns

$ \def\R#1{{\mathbb R}^{#1}} \def\o{{\tt1}} \def\k{\otimes} \def\h{\odot} \def\BR#1{\Big[#1\Big]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\vc#1{\op{vec}\LR{#1}} \def\frob#1{\Big\| ...
greg's user avatar
  • 36.6k
2 votes

Linear matrix equation with stacked matrices and repeated unknowns

Why don't you split up $P$ like this: $$ P = \begin{bmatrix} P_1 \\ P_2 \\ \vdots \\P_n\\ \end{bmatrix}$$ Then you have a set of linear systems $KA_i = P_i$ for each $i = 1, \dots, n$. Here the $A_i$ ...
Lee Fisher's user avatar
  • 2,129
2 votes

How to solve $AX=B$ for $X$ if $A$ is not invertible?

$$A\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}$$ $$\iff\begin{cases}5x+5y-3z&=a\\2x+2y-z&=b\\-x-y+z&=c\end{cases}$$ $$\iff\begin{cases}z&=c+x+y\\x+y-c&=...
Anne Bauval's user avatar
  • 37.7k
1 vote

How to solve $AX=B$ for $X$ if $A$ is not invertible?

Just write down the $9$ linear equations in the $9$ variables of $X$ from the equation $AX=B$. If the first row of $X$ is given by $(a,b,c)$, then the remaining $6$ variables can be expressed by $a,b,...
Dietrich Burde's user avatar
1 vote

Find the determinant of the linear transformation $T(f) = 7f-3f^{'}+7f^{''}$ from the space $V$ spanned by $\cos x$ and $\sin x$ to $V$

Your answer is right, the determinant is $+9$. More ‘conceptually’, let $D:V\to V$ be the differentiation operator $D(f)=f’$. The matrix representation of $D$ relative to the basis $\{\sin, \cos\}$ is ...
peek-a-boo's user avatar
  • 57.4k
1 vote

$ABACA = 0 \Longrightarrow BAC = 0$ if $A,B,C \ge 0$ are symmetric.

This proof looks correct! Indeed, for an matrix $M$, if $M^T M$, then for any vector $x$, $$ \lVert Mx\rVert^2 = x^T M^T M x = 0, $$ from which any $Mx = 0$ and so $M = 0$. In terms of a concise ...
Damian Pavlyshyn's user avatar
1 vote
Accepted

$ABACA = 0 \Longrightarrow BAC = 0$ if $A,B,C \ge 0$ are symmetric.

Note that for real matrices $M$, $M = 0$ if and only if $M^TM = 0$. More generally, for a positive semidefinite matrix $P$, it holds that $M^TPM = 0$ if and only if $\sqrt{P}M = 0$, which holds if and ...
Ben Grossmann's user avatar
1 vote

$ABACA = 0 \Longrightarrow BAC = 0$ if $A,B,C \ge 0$ are symmetric.

Your solution looks essentially correct. Denote the inner product of two vectors $f,g$ by $\langle f,g \rangle.$ For any vector $f$ we have $$ \langle C^{1/2}ABAf,C^{1/2}ABAf\rangle = \langle(ABACA)...
Steen82's user avatar
  • 220
1 vote

Binary matrix power for a specific entry.

Let's look at the formula for increasing $k$, \begin{align} (A^1)_{i, j} &= A_{i, j} \\ (A^2)_{i, j} &= \sum_m A_{i,m} A_{m,j} \\ (A^3)_{i, j} &= \sum_{m} (A^2)_{i, m} A_{m,j} \\ &= ...
Tony Mathew's user avatar
  • 2,407

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